CBSE Class 12 Physics Heat and Thermodynamics Solved Examples

CBSE Class 12 Physics Heat and Thermodynamics Solved Examples. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Ex.1 The temperatures of equal masses of three different liquids A, B and C are 12ºC, 19ºC and 28ºC respectively. The temperature when A and B are mixed is 16ºC and when B and C are mixed, it is 23ºC. Find the temperature of mixture when A and C liquids are mixed.
Sol.: Let m be mass of each liquid and SA, SB and SC be their respective specific heats. Heat lost by B = Heat gained by A

Ex.2 A mixture of 250 gm of water and 200 gm of ice at 0ºC is kept in a calorimeter which has a water equivalent of 50 gm. If 200 gm of steam at 100ºC is passed through this mixture, calculate the final temperature and weight of the contents of the calorimeter. Latent heat of fusion of ice = 80 cal/gm., Latent heat of vaporization of water to steam = 540 cal/gm., Specific heat of water = 1 cal/gm/ºC.
Sol.: Heat lost by 200 gm of steam before it is condensed to water at 100ºC

= 200 × 540 = 108000 cal. .....(i)
Heat gained by 200 gm of ice at 0ºC = mL + m × S × ΔT

= 200 × 80 + 200 × 1 × (100 – 0) = 36000 cal

Heat gained by 250 gm of water and 50 gm of water equivalent of calorimeter at 100ºC to 0ºC
= 250 × 1 × (100 – 0) + 50 × (100 – 0)
= 300 × 100 = 30000 cal.
Total heat gained = 30000 cal. + 36000 = 66000 cal. .....(ii)
Amount of heat lost by steam (i) is greater than heat gained by ice. This shows that only a part of the steam will condense to water at 100ºC which will be sufficient for melting ice.
Let M be mass of steam which will be sufficient to melt ice,
∴ Mass M of steam required is given by M × 540 = 66000
or M = 66000/540 = 9
1100 = 122.2 gm.
Final temperature of system = 100ºC
Weight of contents = weight of ice + water + steam condensed
= 250 + 200 + 122.2
= 572.2 gm.

Ex.3 A copper calorimeter of mass 100 gm contains 200 g of a mixture of ice and water. Steam at 100ºC under normal pressure is passed into the calorimeter and the temperature of the mixture is allowed to rise to 50ºC. If the mass of the calorimeter and its contents is now 330 gm, what was the ratio of ice and water in the beginning ? Neglect heat losses. Given that :
Specific heat of copper = 0.42 × 103 J/kgº Kº.
Specific heat of water = 4.2 × 103 J/kg Kº
Latent heat of fusion of ice = 3.36 × 105 J/kg Kº
Latent heat of condensation of steam = 22.5 × 105 J/kg K.
Sol.: Heat is lost by steam in getting condensed and heat is gained by water, ice and the calorimeter.
Let the calorimeter contain originally x gm of ice and (200 – x) gm of water.

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