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
• B. 80.45 g
• C. 5.25 g
• D. 33.2 g

Answer 1

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 0^oC to water at 0^o = m x 10^-3 x 336 x 10³ = 336m joule

Heat gained by water to change its temperature from 0^oC to 5^oC = 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 30^oC to 5^oC

= 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.