The calorimeter contains 500 g of water and 300 g of ice at 0 ° C. What temperature is set in the calorimeter

The calorimeter contains 500 g of water and 300 g of ice at 0 ° C. What temperature is set in the calorimeter if you add 100 g of boiling water to it? Specific heat capacity of water 4200 J / kg ° C; specific heat of melting of ice 330 kJ / kg

The heat of the boiling water will first be spent on melting the ice, and only after that (if enough) to increase the temperature of the water.
Let’s find out if 100 g of boiling water can melt 300 g of ice.

Heat required to melt ice:
Qп = L * m = 330,000 J / kg * 0.3 kg = 99,000 J (L is the specific heat of melting of ice).

The maximum amount of heat that boiling water can give when cooled from 100 ° С to 0 ° С:
Qp = rMk (Tn – Tk) = 4200 J / (kg * ° C) * 0.1 kg * 100 ° C = 42000 J (r is the specific heat capacity of water).

Conclusion: To melt 300 g of ice, more heat is required than 100 g of boiling water can give, so all the ice will not be melted. The temperature in the calorimeter will remain at 0 ° C.

Answer: The calorimeter will contain a mixture of water and ice. The temperature in the calorimeter is 0 ° С.