# Calorimetry

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Question

# 2 kg of ice at -20°C is mixed with 5 kg of water at 20°C in an insulating vessel having a negligible heat capacity. Calculate the final mass of water in the vessel in kg. It is given that the specific heat of water and ice are 1 kcal/kg/°C and 0.5 kcal/kg/°C respectively and the latent heat of fusion of ice is 80 kcal/kg.

Moderate
Solution

## Let m kg be the mass of ice melted into water. Heat lost by 5 kg of water = 5 kg $×$ 1 kcal/kg/°C $×$ 20°C = 100 kcal. Heat gained = $mL+ms\Delta T$=m kg $×$ 80 kcal/kg + 2 kg $×$ 0.5 kcal/kg/°C $×$ 20°C = 80 m kcal + 20 kcal. Now, heat gained = heat lost. Therefore, 80 m + 20 = 100or m=1 kg. Therefore, final mass of water = 5 kg + 1 kg = 6 kg.

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A calorimeter of water equivalent 10 g contains a liquid of mass 50 g at 40oC. When m gram of ice at -10oC is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20oC. It is known that specific heat capacity of the liquid changes with temperature as where $\mathrm{\theta }$ is temperature in oC. The specific heat capacity of ice, water and the calorimeter remains constant and values are  and latent heat of fusion of ice is ${\mathrm{L}}_{\mathrm{f}}=80{\mathrm{calg}}^{-1}$. Assume no heat loss to the surrounding and calculate the value of m in grams.