Q5 of 41 Page 47

Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates. Most of the energy needed for evaporation is taken from the water itself and the water is cooled down. Assume that a pitcher contains 10 kg of water and 0.2 g of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decreases by 5°C. Specific heat capacity of water = 4200 J kg–1 °C–1 and latent heat of vaporization of water = 2.27 × 106 J kg–1.

Given


Specific heat capacity of water = 4200 J kg–1 °C–1


latent heat of vaporization of water = 2.27 × 106 J kg–1.


Formula used:


ΔQ=MCΔT


ΔQ=heat exchange


ΔT=tempeature change


M=mass


C=heat capacity


Heat lost per second from the atmosphere = Heat gained by the water


0.2× 10-3× 2.27× 106=4200×(10-0.2× 10-3)× ΔT


ΔT=0.01080974° C


Therefore time required for 5° change=5/0.01080974=462.5 seconds=7.7 minutes


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3

The temperatures of equal masses of three different liquids A, B and C are 12°C, 19°C and 28°C respectively. The temperature when A and B are mixed is 16°C, and when B and C are mixed, it is 23°C. What will be the temperature when A and C are mixed?

 4

Four 2 cm × 2 cm × 2 cm cubes of ice are taken out from a refrigerator and are put in 200 ml of a drink at 10°C.

(a) Find the temperature of the drink when thermal equilibrium is attained in it.


(b) If the ice cubes do not melt completely, find the amount melted. Assume that no heat is lost to the outside of the drink and that the container has negligible heat capacity. Density of ice = 900 kg m–3, density of the drink = 4200 J kg–1 K–1, latent heat of fusion of ice = 3.4 × 105 J kg–1.


 6

A cube of iron (density = 8000 kg m–3, specific heat capacity = 470 J kg–1 K–1) is heated to a high temperature and is placed on a large block of ice at 0°C. The cube melts the ice below it, displaces the water and sinks. In the final equilibrium position, its upper surface just goes inside the ice. Calculate the initial temperature of the cube. Neglect any loss of heat outside the ice and the cube. The density of ice = 900 kg m–3 and the latent heat of fusion of ice = 3.36 × 105 J kg–1.

 7

1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 105 J kg–1 and latent heat of vaporization of water = 2.26 × 106 J kg–1.