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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 g


Let 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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