{"id":16059,"date":"2024-03-16T04:26:16","date_gmt":"2024-03-15T22:56:16","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=16059"},"modified":"2024-03-16T12:05:35","modified_gmt":"2024-03-16T06:35:35","slug":"selina-icse-solutions-class-10-physics-calorimetry","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/selina-icse-solutions-class-10-physics-calorimetry\/","title":{"rendered":"Selina Concise Physics Class 10 ICSE Solutions Calorimetry"},"content":{"rendered":"<h2><span style=\"color: #00ccff;\">Selina Concise Physics Class 10 ICSE Solutions Calorimetry<\/span><\/h2>\n<p>APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 11 Calorimetry. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.<\/p>\n<p><a class=\"button-red\" title=\"Formulae Handbook For ICSE Class 9 and 10 Maths and Science\" href=\"https:\/\/goo.gl\/FCvuh4\">Download Formulae Handbook For ICSE Class 9 and 10<\/a><\/p>\n<p><a class=\"button-red\" title=\"ICSE Solutions for Class 10, 9, 8, 7 and 6\" href=\"https:\/\/www.aplustopper.com\/icse-solutions\/\">ICSE Solutions<\/a><a class=\"button-blue\" title=\"Selina Concise ICSE Solutions for Class 10, 9, 8, 7 and 6\" href=\"https:\/\/www.aplustopper.com\/selina-concise-icse-solutions\/\">Selina ICSE Solutions<\/a><\/p>\n<p><strong>Selina ICSE Solutions for Class 10 Physics Chapter 11 Calorimetry<\/strong><\/p>\n<p><span style=\"color: #0000ff;\"><strong>Exercise 11(A)<\/strong><\/span><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 1.<\/strong><\/span><\/p>\n<p>The kinetic energy due to random motion of the molecules of a substance is known as its heat energy.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 2.<\/strong><\/span><\/p>\n<p>S.I. unit of heat is joule (symbol J).<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 3.<\/strong><\/span><\/p>\n<p>One calorie of heat is the heat energy required to raise the temperature of 1 g of water from 14.5<sup>o<\/sup>C to 15.5<sup>\u00a0o<\/sup>C.<br \/>\n1 calorie = 4.186 J<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 4.<\/strong><\/span><\/p>\n<p>One kilo-calorie of heat is the heat energy required to raise the temperature of 1 kg of water from 14.5<sup>o<\/sup>C to 15.5<sup>o<\/sup>C.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 5.<\/strong><\/span><\/p>\n<p>The quantity which determines the direction of flow of heat between two bodies kept in contact is called temperature.<br \/>\nS.I. unit kelvin (K).<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 6.<\/strong><\/span><\/p>\n<table border=\"2\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"260\"><strong>Heat<\/strong><\/td>\n<td style=\"text-align: center;\" width=\"240\"><strong>Temperature<\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"260\">The kinetic energy due to random motion of the molecules of\u00a0a substance is known as its heat energy.<\/td>\n<td style=\"text-align: center;\" width=\"240\">The quantity which determines the direction of flow of heat between two bodies kept in contact is called temperature.<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"260\">S.I. unit joule (J).<\/td>\n<td style=\"text-align: center;\" width=\"240\">S.I. unit kelvin (K).<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"260\">It is measured by the principle of calorimetry.<\/td>\n<td style=\"text-align: center;\" width=\"240\">It is measured by a thermometer.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><span style=\"color: #eb4924;\"><strong>Solution 7.<\/strong><\/span><\/p>\n<p>The measurement of the quantity of heat is called calorimetry.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 8.<\/strong><\/span><\/p>\n<p>The heat capacity of a body is the amount of heat energy required to raise its temperature by 1<sup>o<\/sup>C or 1K.<br \/>\nS.I. unit is joule per kelvin (JK<sup>-1<\/sup>).<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 9.<\/strong><\/span><\/p>\n<p>The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of unit mass of that substance through by 1<sup>o<\/sup>C (or 1K).<br \/>\nS.I. unit is joule per kilogram per kelvin (Jkg<sup>-1<\/sup>K<sup>-1<\/sup>).<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 10.<\/strong><\/span><\/p>\n<p>Heat capacity = Mass x specific heat capacity<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 11.<\/strong><\/span><\/p>\n<ul>\n<li>Heat capacity of the body is the amount of heat required to raise the temperature of (whole) body by 1<sup>o<\/sup>C\u00a0whereas specific heat capacity is the amount of heat required to raise the temperature of unit mass of the body by 1<sup>o<\/sup>C.<\/li>\n<li>Heat capacity of a substance depends upon the material and mass of the body. Specific heat capacity of a substance does not depend on the mass of the body.<\/li>\n<li>S.I. unit of heat capacity is JK<sup>-1<\/sup>\u00a0and S.I. unit of specific heat capacity is Jkg<sup>-1<\/sup>K<sup>-1<\/sup>.<\/li>\n<\/ul>\n<p><span style=\"color: #eb4924;\"><strong>Solution 12.<\/strong><\/span><\/p>\n<p>Water has the highest specific heat capacity.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 13.<\/strong><\/span><\/p>\n<p>Specific heat capacity of water=4200 J kg<sup>-1\u00a0<\/sup>K<sup>-1<\/sup>.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 14.<\/strong><\/span><\/p>\n<ol>\n<li>The heat capacity of a body is 50JK<sup>-1<\/sup>\u00a0means to increase the temperature of this body by 1K we have to supply 50 joules of energy.<\/li>\n<li>The specific heat capacity of copper is 0.4Jg<sup>-1<\/sup>K<sup>-1<\/sup>\u00a0means to increase the temperature of one gram of copper by 1K we have to supply 0.4 joules of energy.<\/li>\n<\/ol>\n<p><span style=\"color: #eb4924;\"><strong>Solution 16.<\/strong><\/span><\/p>\n<p>The quantity of heat energy absorbed by a body depends on three factors :<\/p>\n<ol>\n<li>Mass of the body &#8211; The amount of heat energy required is directly proportional to the mass of the substance.<\/li>\n<li>Nature of material of the body &#8211; The amount of heat energy required depends on the nature on the substance and it is expressed in terms of its specific heat capacity c.<\/li>\n<li>Rise in temperature of the body &#8211; The amount of heat energy required is directly proportional to the rise in temperature.<\/li>\n<\/ol>\n<p><span style=\"color: #eb4924;\"><strong>Solution 17.<\/strong><\/span><\/p>\n<p>The expression for the heat energy Q<br \/>\nQ = mc\u0394t\u00a0(in joule)<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 18.<\/strong><\/span><\/p>\n<p>Heat capacity of liquid A is less than that of B.<br \/>\nAs the substance with low heat capacity shows greater rise in temperature.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 19.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92072\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-1.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 1\" width=\"488\" height=\"165\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-1.png 488w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-1-300x101.png 300w\" sizes=\"(max-width: 488px) 100vw, 488px\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 20.<\/strong><\/span><\/p>\n<p>The principle of method of mixture:<br \/>\nHeat energy lost by the hot body = Heat energy gained by the cold body.<br \/>\nThis principle is based on law of conservation of energy.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 21.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92073\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-2.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 2\" width=\"430\" height=\"253\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-2.png 430w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-2-300x177.png 300w\" sizes=\"(max-width: 430px) 100vw, 430px\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 22.<\/strong><\/span><\/p>\n<p>In the absence of water, if on a cold winter night the atmospheric temperature falls below 0<sup>o<\/sup>C, the water in the fine capillaries of plant will freeze, so the veins will burst due to the increase in the volume of water on freezing. As a result, plants will die and the crop will be destroyed. In order to save the crop on such cold nights, farmers fill their fields with water because water has high specific heat capacity, so it does not allow the temperature in the surrounding area of plants to fall up to 0<sup>o<\/sup>C.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 23.<\/strong><\/span><\/p>\n<p>The specific heat capacity of water is very high. It is about five times as high as that of sand. Hence the heat energy required for the same rise in temperature by a certain mass of water will be nearly five times than that required by the same mass of sand. Similarly, a certain mass of water will give out nearly five times more heat energy than that given by sand of the same mass for the same fall in temperature. As such, sand gets heated or cooled more rapidly as compared to water under the similar conditions. Thus a large difference in temperature is developed between the land and the sea due to which land and sea breezes are formed. These breezes make the climate near the sea shore moderate.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 24.<\/strong><\/span><\/p>\n<p>The reason is that water does not cool quickly due to its large specific heat capacity, so a hot water bottle provides heat energy for fomentation for a long time.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 25.<\/strong><\/span><\/p>\n<p>By allowing water to flow in pipes around the heated parts of a machine, heat energy from such part is removed. Water in pipes extracts more heat from surrounding without much rise in its temperature because of its large specific heat capacity. So, Water is used as an effective coolant.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 26.<\/strong><\/span><\/p>\n<ol>\n<li>Radiator in car.<\/li>\n<li>To avoid freezing of wine and juice bottles.<\/li>\n<\/ol>\n<p><span style=\"color: #eb4924;\"><strong>Solution 28.<\/strong><\/span><\/p>\n<p>A calorimeter is a cylindrical vessel which is used to measure the amount of heat gained or lost by a body when it is mixed with other body.<br \/>\nIt is made up of thin copper sheet because:<\/p>\n<ol>\n<li>Copper is a good conductor of heat, so the vessel soon acquires the temperature of its contents.<\/li>\n<li>Copper has low specific heat capacity so the heat capacity of calorimeter is low and the amount of heat energy taken by the calorimeter from its contents to acquire the temperature of its contents is negligible.<\/li>\n<\/ol>\n<p><span style=\"color: #eb4924;\"><strong>Solution 29.<\/strong><\/span><\/p>\n<p>By making the base of a cooking pan thick, its thermal capacity becomes large and it imparts sufficient heat energy at a low temperature to the food for its proper cooking. Further it keeps the food warm for a long time, after cooking.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 1 (MCQ).<\/strong><\/span><\/p>\n<p>JK<sup>-1<\/sup><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 2 (MCQ).<\/strong><\/span><\/p>\n<p>J kg<sup>-1<\/sup>K<sup>-1<\/sup><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 3 (MCQ).<\/strong><\/span><\/p>\n<p>4200 J kg<sup>-1<\/sup>K<sup>-1<\/sup><\/p>\n<p><span style=\"color: #0000ff;\"><strong>Numericals<\/strong><\/span><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 1.<\/strong><\/span><\/p>\n<p>The size of 1 degree on the Kelvin scale is the same as the size of 1 degree on the Celsius scale. Thus, the difference (or change) in temperature is the same on both the Celsius and Kelvin scales.<br \/>\nTherefore, the corresponding rise in temperature on the Kelvin scale will be 15K.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 2.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92074\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-3.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 3\" width=\"357\" height=\"360\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-3.png 357w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-3-150x150.png 150w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-3-298x300.png 298w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-3-100x100.png 100w\" sizes=\"(max-width: 357px) 100vw, 357px\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 3.<\/strong><\/span><\/p>\n<ol>\n<li>We know that heat energy needed to raise the temperature by 15<sup>o<\/sup>\u00a0is = heat capacity x change in temperature.<br \/>\nHeat energy required= 966 J K<sup>-1<\/sup>\u00a0x 15 K = 14490 J.<\/li>\n<li>We know that specific heat capacity is = heat capacity\/ mass of substance<br \/>\nSo specific heat capacity is = 966 \/ 2=483 J kg<sup>-1\u00a0<\/sup>K<sup>-1<\/sup>.<\/li>\n<\/ol>\n<p><span style=\"color: #eb4924;\"><strong>Solution 4.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92075\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-4.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 4\" width=\"442\" height=\"237\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-4.png 442w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-4-300x161.png 300w\" sizes=\"(max-width: 442px) 100vw, 442px\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 5.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92077\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-5.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 5\" width=\"340\" height=\"267\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-5.png 340w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-5-300x236.png 300w\" sizes=\"(max-width: 340px) 100vw, 340px\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 6.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92078\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-6.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 6\" width=\"372\" height=\"383\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-6.png 372w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-6-291x300.png 291w\" sizes=\"(max-width: 372px) 100vw, 372px\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 7.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92080\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-7.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 7\" width=\"362\" height=\"384\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-7.png 362w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-7-283x300.png 283w\" sizes=\"(max-width: 362px) 100vw, 362px\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 8.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92081\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-8.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 8\" width=\"251\" height=\"195\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 9.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92082\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-9.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 9\" width=\"399\" height=\"459\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-9.png 399w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-9-261x300.png 261w\" sizes=\"(max-width: 399px) 100vw, 399px\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 10.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92083\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-10.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 10\" width=\"372\" height=\"453\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-10.png 372w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-10-246x300.png 246w\" sizes=\"(max-width: 372px) 100vw, 372px\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 11.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92084\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-11.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 11\" width=\"276\" height=\"384\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-11.png 276w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-11-216x300.png 216w\" sizes=\"(max-width: 276px) 100vw, 276px\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 12.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92085\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-12.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 12\" width=\"340\" height=\"418\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-12.png 340w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-12-244x300.png 244w\" sizes=\"(max-width: 340px) 100vw, 340px\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 13.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92086\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-13.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 13\" width=\"510\" height=\"124\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-13.png 510w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-13-300x73.png 300w\" sizes=\"(max-width: 510px) 100vw, 510px\" \/><\/p>\n<p><span style=\"color: #0000ff;\"><strong>Exercise 11(B)<\/strong><\/span><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 1.<\/strong><\/span><\/p>\n<p>(a) The process of change from one state to another at a constant temperature is called the change of phase of substance.<br \/>\n(b) There is no change in temperature during the change of phase.<br \/>\n(c) Yes, the substance absorbs or liberates heat during the change of phase.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 2.<\/strong><\/span><\/p>\n<p><strong>Melting:<\/strong> The change from solid to liquid phase on heating at a constant temperature is called melting.<br \/>\n<strong>Melting point:<\/strong> The constant temperature at which a solid changes to liquid is called the melting point.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 3.<\/strong><\/span><\/p>\n<ol>\n<li>Average kinetic energy of molecules changes.<\/li>\n<li>Average potential energy of molecules changes.<\/li>\n<\/ol>\n<p><span style=\"color: #eb4924;\"><strong>Solution 4.<\/strong><\/span><\/p>\n<p>(a) Average kinetic energy does not change.<br \/>\n(b) Average potential energy increases.<br \/>\n<strong>Explanation:<\/strong> When a substance is heated at constant temperature (i.e. during its phase change state), the heat supplied makes the vibrating molecules gain potential energy to overcome the intermolecular force of attraction and move about freely. This means that the substance changes its form.<\/p>\n<p>However, this heat does not increase the kinetic energy of the molecules, and hence, no rise in temperature takes place during the change in phase of a substance.<br \/>\nThis heat supplied to the substance is known as latent heat and is utilized in changing the state of matter without any rise in temperature.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 5.<\/strong><\/span><\/p>\n<p>The melting point of ice decreases by the presence of impurity in it.<br \/>\nUse: In making the freezing mixture by adding salt to ice. This freezing mixture is used in preparation of ice creams.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 6.<\/strong><\/span><\/p>\n<p>The melting point of ice decreases by the increase in pressure. The melting point of ice decrease by 0.0072<sup>o<\/sup>C for every one atmosphere rise in pressure.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 7.<\/strong><\/span><\/p>\n<p>(a) AB part shows rise in temperature of solid from 0 to T<sub>1<\/sub><sup>o<\/sup>C, BC part shows melting at temperature T<sub>1<\/sub><sup>o<\/sup>C, CD part shows rise in temperature of liquid from T<sub>1<\/sub><sup>o<\/sup>C to T<sub>3<\/sub><sup>o<\/sup>C , DE part shows the boiling at temperature T<sub>3<\/sub><sup>o<\/sup>C.<br \/>\n(b) T<sub>1<\/sub><sup>o<\/sup>C.<br \/>\n(c) T<sub>3<\/sub><sup>o<\/sup>C.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 8.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92087\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-14.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 14\" width=\"418\" height=\"256\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-14.png 418w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-14-300x184.png 300w\" sizes=\"(max-width: 418px) 100vw, 418px\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 9.<\/strong><\/span><\/p>\n<p><strong>Boiling:<\/strong> The change from liquid to gaseous phase on heating at a constant temperature is called boiling.<br \/>\n<strong>Boiling point:<\/strong> The particular temperature at which vaporization occurs is called the boiling point of liquid.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 10.<\/strong><\/span><\/p>\n<p>Volume of water wills increases when it boils at 100<sup>o<\/sup>C.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 11.<\/strong><\/span><\/p>\n<p>The boiling point of water increases on adding salt.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 12.<\/strong><\/span><\/p>\n<p>The boiling point of a liquid increases on increasing the pressure.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 14.<\/strong><\/span><\/p>\n<p>In a pressure cooker, the water boils at about 120<sup>o<\/sup>C to 125<sup>o<\/sup>C.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 15.<\/strong><\/span><\/p>\n<p>This is because at high altitudes atmospheric pressure is low; therefore boiling point of water decreases and so it does not provide the required heat energy for cooking.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 16.<\/strong><\/span><\/p>\n<p>(a) When ice melts, its volume <strong>decreases.<\/strong><br \/>\n(b) Decrease in pressure over ice<strong> increases<\/strong> its meltingpoint.<br \/>\n(c) Increase in pressure <strong>increases<\/strong> the boiling point of water.<br \/>\n(d)A pressure cooker is based on the principle that boiling point of water increases with the <strong>increase in pressure.<\/strong><br \/>\n(e) The boiling point of water is defined as <strong>the constant temperature at which water changes to steam.<\/strong><br \/>\n(f) water can be made to boil at 115\u00b0C by <strong>increasing<\/strong> pressure over its surface.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 17.<\/strong><\/span><\/p>\n<p>Latent heat: The heat energy exchanged in change of phase is not externally manifested by any rise or fall in temperature, it is considered to be hidden in the substance and is called the latent heat.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 18.<\/strong><\/span><\/p>\n<p>The quantity of heat required to convert unit mass of ice into liquid water at 0<sup>0<\/sup>C (melting point) is called the specific latent heat of fusion of ice.<br \/>\nIts S.I. unit is Jkg<sup>-1<\/sup>.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 19.<\/strong><\/span><\/p>\n<p>Specific latent heat of ice: 336000 J kg<sup>-1<\/sup>.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 20.<\/strong><\/span><\/p>\n<p>It means 1 g of ice at 0<sup>o<\/sup>C absorbs 360 J of heat energy to convert into water at 0<sup>o<\/sup>C.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 21.<\/strong><\/span><\/p>\n<p>Latent heat is absorbed by ice.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 22.<\/strong><\/span><\/p>\n<p>1 g of water at 0<sup>o<\/sup>C has more heat than 1 g of ice at 0<sup>o<\/sup>C. This is because ice at 0<sup>o<\/sup>C absorbs 360 J of heat energy to convert into water at 0<sup>o<\/sup>C.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 23.<\/strong><\/span><\/p>\n<p>(a) 1 g ice at 0<sup>o<\/sup>C requires more heat because ice would require additional heat energy equal to latent heat of melting.<br \/>\n(b) 1 g ice at 0<sup>o<\/sup>C first absorbs 336 J heat to convert into 1 g water at 0<sup>o<\/sup>C.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 24.<\/strong><\/span><\/p>\n<p>This is because 1 g of ice at 0<sup>o<\/sup>C takes 336 J of heat energy from the mouth to melt at 0<sup>o<\/sup>C. Thus mouth loses an additional 336 J of heat energy for 1 g of ice at 0<sup>o<\/sup>C than for 1g of water at 0<sup>o<\/sup>C. Therefore cooling produced by 1 g of ice at 0<sup>o<\/sup>C is more than for 1g of water at 0<sup>o<\/sup>C.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 25.<\/strong><\/span><\/p>\n<p>This is because 1 g of ice at 0<sup>o<\/sup>C takes 336 J of heat energy from the bottle to melt into water at 0<sup>o<\/sup>C. Thus bottle loses an additional 336 J of heat energy for 1 g of ice at 0<sup>o<\/sup>C than for 1 g iced water at 0<sup>o<\/sup>C. Therefore bottled soft drinks get cooled, more quickly by the ice cubes than by iced water.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 26.<\/strong><\/span><\/p>\n<p>The reason is that after the hail storm, the ice absorbs the heat energy required for melting from the surrounding, so the temperature of the surrounding falls further down and we feel colder.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 27.<\/strong><\/span><\/p>\n<p>The reason is that the heat energy required for melting the frozen lake is absorbed from the surrounding atmosphere. As a result, the temperature of the surrounding falls and it became very cold.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 29.<\/strong><\/span><\/p>\n<ol>\n<li>The reason is that the specific latent heat of fusion of ice is sufficiently high, so when the water of lake freezes, a large quantity of heat has to be released and hence the surrounding temperature becomes pleasantly warm.<\/li>\n<li>Heat supplied to a substance during its change of state, does not cause any rise in its temperature because this is latent heat of phase change which is required to change the phase only.<\/li>\n<\/ol>\n<p><span style=\"color: #eb4924;\"><strong>Solution 1 (MCQ).<\/strong><\/span><\/p>\n<p>J kg<sup>-1<\/sup><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 2 (MCQ).<\/strong><\/span><\/p>\n<p>80cal g<sup>-1<\/sup><\/p>\n<p><span style=\"color: #0000ff;\"><strong>Numericals<\/strong><\/span><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 1.<\/strong><\/span><\/p>\n<p>Mass of ice=10g = 0.01kg<br \/>\nAmount of heat energy absorbed, Q=5460J<br \/>\nSpecific latent heat of fusion of ice=?<br \/>\nSpecific heat capacity of water = 4200Jkg<sup>-1<\/sup>K<sup>-1<br \/>\n<\/sup>Amount of heat energy required by 10g (0.01kg) of water at 0<sup>o<\/sup>C to raise its temperature by 50<sup>o<\/sup>C= 0.01X4200X50=2100J.<br \/>\nLet Specific latent heat of fusion of ice=L Jg<sup>-1<\/sup>.<br \/>\nThen,<br \/>\nQ = mL + mc\u0394T<br \/>\n5460 J = 10 x L + 2100J<br \/>\nL = 336Jg<sup>-1<\/sup>.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 2.<\/strong><\/span><\/p>\n<p>Mass of water m = 5.0 g<br \/>\nspecific heat capacity of water c = 4.2 J g<sup>-1\u00a0<\/sup>K<sup>-1<br \/>\n<\/sup>specific latent heat of fusion of iceL =336 J g<sup>-1<br \/>\n<\/sup>Amount of heat energy released when 5.0 g of water at 20<sup>o<\/sup>C changes into water at 0<sup>o<\/sup>C = 5 x 4.2 x 20 = 420 J.<br \/>\nAmount of heat energy released when 5.0g of water at 0<sup>o<\/sup>C changes into ice at 0<sup>o<\/sup>C=5x336J=1680J.<br \/>\nTotal amount of heat released =1680 J + 420 J = 2100 J.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 3.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92088\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-15.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 15\" width=\"341\" height=\"217\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-15.png 341w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-15-300x191.png 300w\" sizes=\"(max-width: 341px) 100vw, 341px\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 4.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92089\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-16.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 16\" width=\"558\" height=\"247\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-16.png 558w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-16-300x133.png 300w\" sizes=\"(max-width: 558px) 100vw, 558px\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 5.<br \/>\n<\/strong><\/span><\/p>\n<p>Mass of ice m<sub>1<\/sub>\u00a0=17 g<br \/>\nMass of water m<sub>2<\/sub>\u00a0=40 g.<br \/>\nChange in temperature =34-0=34K<br \/>\nSpecific heat capacity of water is 4.2Jg<sup>-1<\/sup>K<sup>-1<\/sup>.<br \/>\nAssuming there is no loss of heat, heat energy gained by ice (latent heat of ice), Q= heat energy released by water<br \/>\nQ = 40 x 34 x 4.2 = 5712 J.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92090\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-17.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 17\" width=\"352\" height=\"45\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-17.png 352w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-17-300x38.png 300w\" sizes=\"(max-width: 352px) 100vw, 352px\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 6.<\/strong><\/span><\/p>\n<p>Let whole of the ice melts and let the final temperature of the mixture be T<sup>o<\/sup>C.<br \/>\nAmount of heat energy gained by 10g of ice at -10<sup>o<\/sup>C to raise its temperature to 0<sup>o<\/sup>C= 10x10x2.1=210J<br \/>\nAmount of heat energy gained by 10g of ice at 0<sup>o<\/sup>C to convert into water at 0<sup>o<\/sup>C=10&#215;336=3360 J<br \/>\nAmount of heat energy gained by 10g of water (obtained from ice) at 0<sup>o<\/sup>C to raise its temperature to T<sup>o<\/sup>C = 10&#215;4.2x(T-0)=42T<br \/>\nAmount of heat energy released by 10g of water at 10<sup>o<\/sup>C to lower its temperature to T<sup>o<\/sup>C = 10&#215;4.2x(10-T)=420-42T<br \/>\nHeat energy gained = Heat energy lost<br \/>\n210 + 3360 + 42T = 420-42T<br \/>\nT = -37.5<sup>o<\/sup>C<br \/>\nThis cannot be true because water cannot exist at this temperature.<br \/>\nSo whole of the ice does not melt. Let m gm of ice melts. The final temperature of the mixture becomes 0<sup>o<\/sup>C.<br \/>\nSo, amount of heat energy gained by 10g of ice at -10<sup>o<\/sup>C to raise its temperature to 0<sup>o<\/sup>C= 10x10x2.1=210J<br \/>\nAmount of heat energy gained by m gm of ice at 0<sup>o<\/sup>C to convert into water at 0<sup>o<\/sup>C=mx336=336m J<br \/>\nAmount of heat energy released by 10g of water at 10<sup>o<\/sup>C to lower its temperature to 0<sup>o<\/sup>C = 10&#215;4.2x(10-0)=420<br \/>\nHeat energy gained = Heat energy lost<br \/>\n210 + 336m = 420<br \/>\nm = 0.625 gm<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 7.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92091\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-18.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 18\" width=\"613\" height=\"361\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-18.png 613w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-18-300x177.png 300w\" sizes=\"(max-width: 613px) 100vw, 613px\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 8.<\/strong><\/span><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92092\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-19.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 19\" width=\"566\" height=\"556\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-19.png 566w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-19-300x295.png 300w\" sizes=\"(max-width: 566px) 100vw, 566px\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 9.<\/strong><\/span><\/p>\n<p>Since the whole block does not melt and only 2 kg of it melts, so the final temperature would be 0<sup>\u00a0o<\/sup>C.<br \/>\nAmount of heat energy gained by 2 kg of ice at 0<sup>o<\/sup>C to convert into water at 0<sup>o<\/sup>C=2 x 336000 = 672000 J<br \/>\nLet amount of water poured = m kg.<br \/>\nInitial temperature of water = 100<sup>o<\/sup>C.<br \/>\nFinal temperature of water = 0<sup>o<\/sup>C.<br \/>\nAmount of heat energy lost by m kg of water at 100<sup>o<\/sup>C to reach temperature 0<sup>o<\/sup>C =m x 4200 x 100 = 420000m J<br \/>\nWe know that heat energy gained = heat energy lost.<br \/>\n672000J = mX420000J<br \/>\nm = 672000\/420000=1.6kg<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 10.<\/strong><\/span><\/p>\n<p>Amount of heat energy gained by 100 g of ice at -10<sup>o\u00a0<\/sup>C to raise its temperature to 0<sup>o\u00a0<\/sup>C = 100 \u00d7\u00a02.1 \u00d7\u00a010 = 2100 J<br \/>\nAmount of heat energy gained by 100 g of ice at 0<sup>o\u00a0<\/sup>C to convert into water at 0<sup>o\u00a0<\/sup>C = 100 \u00d7\u00a0336 = 33600 J<br \/>\nAmount of heat energy gained when temperature of 100 g of water at 0<sup>o\u00a0<\/sup>C rises to 100<sup>o\u00a0<\/sup>C = 100 \u00d7\u00a04.2 \u00d7\u00a0100 = 42000 J<br \/>\nTotal amount of heat energy gained is = 2100 + 33600 + 42000 = 77700 J = 7.77\u00a0\u00d7 10<sup>4<\/sup>\u00a0J<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 11.<\/strong><\/span><\/p>\n<p>Amount of heat energy gained by 1kg of ice at -10<sup>o<\/sup>C to raise its temperature to 0<sup>o<\/sup>C= 1 x 2100 x 10 = 21000 J<br \/>\nAmount of heat energy gained by 1kg of ice at 0<sup>o<\/sup>C to convert into water at 0<sup>o<\/sup>C = L<br \/>\nAmount of heat energy gained when temperature of 1kg of water at 0<sup>o<\/sup>C rises to 100<sup>o<\/sup>C= 1 x 4200 x 100 = 420000 J<br \/>\nTotal amount of heat energy gained = 21000 + 420000 + L = 441000 + L.<br \/>\nGiven that total amount of heat gained is = 777000J.<br \/>\nSo,<br \/>\n441000 + L = 777000.<br \/>\nL = 777000 &#8211; 441000.<br \/>\nL = 336000JKg<sup>-1<\/sup><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-92093\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/11\/Selina-Concise-Physics-Class-10-ICSE-Solutions-Calorimetry-img-20.png\" alt=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry img 20\" width=\"261\" height=\"299\" \/><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 12.<\/strong><\/span><\/p>\n<p>Mass of ice, mice = 200 g<\/p>\n<p>Time for ice to melt, t1 = 1 min = 60 s<\/p>\n<p>Mass of water, mw = 200 g<\/p>\n<p>Temperature change of water,\u00a0\u0394T = 20\u00a0\u00b0C<\/p>\n<p>Rate of heat exchange is constant. So, power required for converting ice to water is same as the power required to increase the temperature of water.<\/p>\n<p><span style=\"color: #0000ff;\"><strong>Exercise 11(C)<\/strong><\/span><\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 1.<\/strong><\/span><\/p>\n<p>The earth receives heat radiations from sun which reach us after passing through its atmosphere. The earth&#8217;s atmosphere is transparent for the visible and thermal radiations of short wavelengths coming from the sun. The earth&#8217;s surface and objects on it thus become warm in the day time. After the sunset, the earth&#8217;s surface and the objects on it radiate the infrared radiations of long wavelengths. A part of these radiations are reflected back by the clouds and a part of it is absorbed by the green house gases like carbon dioxide, methane, water vapours and chlorofluorocarbons. Thus the clouds and green house gases prevents a large fraction of radiations given out by the earth&#8217;s surface, from escaping into the space. This phenomenon is called greenhouse effect.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 2.<\/strong><\/span><\/p>\n<p>Carbon dioxide, methane, water vapours and chlorofluorocarbons.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 3.<\/strong><\/span><\/p>\n<p>(a) Short wavelength radiations<br \/>\n(b) Long wavelength radiations<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 4.<\/strong><\/span><\/p>\n<p>Coal, petroleum, natural gas.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 5.<\/strong><\/span><\/p>\n<p>From sun, we receive 1366W m<sup>-2<\/sup>\u00a0energy at the top of our earth&#8217;s atmosphere, out of which only 235Wm<sup>-2\u00a0<\/sup>energy reaches near the earth&#8217;s surface. The earth and ocean surface absorbs 168 W m<sup>-2<\/sup>\u00a0energy and only 67 W m<sup>-2<\/sup>\u00a0energy remains in the lower atmosphere. With this much energy received on earth surface, its actual surface temperature would have been around -18<sup>0<\/sup>C which is quite uncomfortable for human living. Fortunately the greenhouse gases present in the earth&#8217;s atmosphere contribute in trapping the heat energy within the atmosphere and they produce an average warming effect of about 33<sup>0<\/sup>C to keep the effective temperature around 15<sup>0<\/sup>C. So, greenhouse effect helps in keeping the temperature of earth&#8217;s surface suitable for living of human beings.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 6.<\/strong><\/span><\/p>\n<p>Three reasons for increase of greenhouse gases:<\/p>\n<ol>\n<li>The burning of fuels, deforestation and industrial production<\/li>\n<li>Increase of population.<\/li>\n<li>Imbalance of carbon dioxide cycle<\/li>\n<\/ol>\n<p><span style=\"color: #eb4924;\"><strong>Solution 7.<\/strong><\/span><\/p>\n<p>The effect of enhancement of greenhouse effect are:<\/p>\n<ol>\n<li>The variable change in the climate in different parts of the world has created difficulty and forced the people and animals to migrate from one place to another place.<\/li>\n<li>It has affected the blooming season of the different plants.<\/li>\n<li>The climate changes have shown the immediate effect on simple organism and plants.<\/li>\n<li>It has affected the world&#8217;s ecology.<\/li>\n<li>It has increased the heat stroke deaths.<\/li>\n<\/ol>\n<p><span style=\"color: #eb4924;\"><strong>Solution 8.<\/strong><\/span><\/p>\n<p>Global warming means the increase in the average effective temperature of earth&#8217;s surface due to an increase in the amount of greenhouse gases in its atmosphere.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 9.<\/strong><\/span><\/p>\n<p>The effect of enhancement of greenhouse effect are:<\/p>\n<ol>\n<li>The variable change in the climate in different parts of the world has created difficulty and forced the people and animals to migrate from one place to another place.<\/li>\n<li>It has affected the blooming season of the different plants.<\/li>\n<li>The climate changes have shown the immediate effect on simple organism and plants.<\/li>\n<li>It has affected the world&#8217;s ecology.<\/li>\n<li>It has increased the heat stroke deaths.<\/li>\n<\/ol>\n<p><span style=\"color: #eb4924;\"><strong>Solution 10.<\/strong><\/span><\/p>\n<p>Due to rise in sea level the building and roads in the coastal areas will get flooded and they could suffer damage from hurricanes and tropical storms.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 11.<\/strong><\/span><\/p>\n<p>Due to global warming, many new diseases have emerged because bacteria can survive better in increased temperature and they can multiply faster. It is extending the distribution of mosquitoes due to increase in humanity levels and their frequent growth in warmer atmosphere. This has resulted in increase of many new diseases. The deaths due to heat stroke have certainly increased.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 12.<\/strong><\/span><\/p>\n<p>Due to global warming, many new diseases have emerged because bacteria can survive better in increased temperature and they can multiply faster. It is extending the distribution of mosquitoes due to increase in humanity levels and their frequent growth in warmer atmosphere. This has resulted in increase of many new diseases. The deaths due to heat stroke have certainly increased.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 13.<\/strong><\/span><\/p>\n<p>At the present rate of increase of green house effect, it is expected that nearly 30% of the plant species will extinct by the year 2050 and up to 70% by the end of the year 2100. In the near future, warming of nearly 3<sup>o<\/sup>C will result in poor yield in farms in low latitude regions. This will increase the rise of malnutrition.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 14.<\/strong><\/span><\/p>\n<p>At the present rate of increase in green house effect, is estimated that nearly 30% of the plant and animal species will extinct by the year 2050 and upto 70 % by the end of year 2100. This will disrupt ecosystem. The animals from the equatorial region will shift to higher latitude in search of cold regions. The absorption of carbon dioxide by ocean will cause acidification due to which marine species will migrate.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 16.<\/strong><\/span><\/p>\n<p>The tax calculated on the basis of: carbon emission from industry, number of employee hour and turnover of the factory is called carbon tax.<br \/>\nThis tax shall be paid by industries. This will encourage the industries to use the energy efficient techniques.<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 1 (MCQ).<\/strong><\/span><\/p>\n<p>-18<sup>o<\/sup>C<\/p>\n<p><span style=\"color: #eb4924;\"><strong>Solution 2 (MCQ).<\/strong><\/span><\/p>\n<p>The increase in sea levels.<br \/>\nExplanation: Due to global warming, the average temperature of the Earth has increased and has lead to the melting of ice around both the poles. This melting of ice has lead to an increase in the level of water in sea.<\/p>\n<p><strong>More Resources for Selina Concise Class 10 ICSE Solutions<\/strong><\/p>\n<ul>\n<li><a href=\"https:\/\/www.aplustopper.com\/selina-icse-solutions-class-10-physics\/\">Selina ICSE Concise Physics Class 10\u00a0Solutions<\/a><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/selina-icse-solutions-class-10-chemistry\/\">Selina Concise Chemistry Class 10 ICSE Solutions<\/a><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/selina-icse-solutions-class-10-biology\/\">Selina Concise Biology Class 10 ICSE Solutions<\/a><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/selina-icse-solutions-class-10-maths\/\">Selina Class 10 ICSE Maths Solutions<\/a><\/li>\n<li><a href=\"https:\/\/paisaalgo.com\/coforge-pivots-calculator\/\">COFORGE Pivot Point Calculator<\/a><\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Selina Concise Physics Class 10 ICSE Solutions Calorimetry APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 11 Calorimetry. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are [&hellip;]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[3034],"tags":[6525,6474,6283,6472,6523,6524],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.8 (Yoast SEO v17.8) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Selina Concise Physics Class 10 ICSE Solutions Calorimetry - A Plus Topper<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aplustopper.com\/selina-icse-solutions-class-10-physics-calorimetry\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry\" \/>\n<meta property=\"og:description\" content=\"Selina Concise Physics Class 10 ICSE Solutions Calorimetry APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 11 Calorimetry. 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