In the calorimeter, which contained 2.5 liters of water with a temperature of 5 degrees C, 800 g of ice was put.

In the calorimeter, which contained 2.5 liters of water with a temperature of 5 degrees C, 800 g of ice was put. When the water temperature stopped changing, it turned out that the ice became 84 g more. Determine the starting ice temperature. Heat exchange with the environment can be neglected. Specific heat of ice – 2.1 kJ / (kg · 0С), water – 4.2 kJ / (kg · 0С), specific heat of melting of ice – 330 kJ / kg.

Task data: mw (initial mass of water) = 2.5 kg (V = 2.5 l); tv (initial water temperature) = 5 ºС; ml1 (initial ice mass) = 800 g = 0.8 kg; ml2 (mass of the formed ice) = 84 g = 0.084 kg; t (equilibrium temperature) = 0 ºС.

Constants: according to the condition Сl (specific heat capacity of ice) = 2.1 * 10 ^ 3 J / (kg * ºС); Sv (specific heat capacity of water) = 4.2 * 10 ^ 3 J / (kg * ºС); λl (specific heat of melting of ice) = 330 * 10 ^ 3 J / kg.

To calculate the initial ice temperature, we use the equality: Сl * ml1 * (t – tl) = Sv * mw * (tw – t) + λl * ml2, from which we can express: tl = t – (Sv * mw * (tw – t) + λl * ml2) / (Сl * ml1).

Calculation: tl = 0 – (4.2 * 10 ^ 3 * 2.5 * (5 – 0) + 330 * 10 ^ 3 * 0.084) / (2.1 * 10 ^ 3 * 0.8) = -47 , 75 ºС.

Answer: The ice had a temperature of -47.75 ºС.