Question

# 250 g of water at 30 °C is present in a copper vessel of mass 50 g. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to 5 °C. Specific latent heat of fusion of ice = 336 x 103 J kg-1 Specific heat capacity of copper vessel= 400 J kg-1 °C-1 Specific heat capacity of water 4200 J kg-1 °C-1

A

74.93 g

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B

80.45 g

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C

5.25 g

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D

33.2 g

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Solution

## The correct option is A 74.93 gLet the required mass of ice be m g. Then heat gained or required to change ice at 0oC to water at 0o = m x 10-3 x 336 x 103 = 336m joule Heat gained by water to change its temperature from 0oC to 5oC = m x 10-3 x 4.2 x 103 x (30 - 5) = 21m joule Total heat gained = (336m + 21m) = 357m ...(i) Now heat lost by water to change its temperature from 30oC to 5oC = 250 x 10-3 x 4.2 x 103 x (30 - 5) = 26,250 joule Heat lost by copper vessel = (50 x 10-3 x 0.4 x 103 x 25) = 500 joule Total heat lost = (26,250 + 500) J = 26,750 J ...(ii) According to the principle of calorimetry, Heat gained = Heat lost 357 m= 26,750 [From (i) and (ii)] m= 26750 / 357 = 74.93g Hence, the required mass of ice is 74.93 g.

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