An aluminum calorimeter weighing 50 g is filled with 120 g of water and an electric heater with a power

An aluminum calorimeter weighing 50 g is filled with 120 g of water and an electric heater with a power of 12.5 W is lowered. How many degrees will the calorimeter with water heat up in 22 minutes if the heat loss to the environment is 20%?

Data: m1 (calorimeter) = 50 g (0.05 kg); m2 (water) = 120 g (0.12 kg); N (heater power) = 12.5 W; t (duration of work) = 22 min (1320 s); η1 (heat loss) = 20% (0.2).

Constants: C1 (specific heat capacity of aluminum) = 920 J / (kg * ºС); C2 (specific heat capacity of water) = 4200 J / (kg * ºС).

Change in temperature of water and calorimeter (Δt): η = 1 – η1 = (C1 * m1 + C2 * m2) * Δt / (N * t); Δt = N * t * (1 – η1) / (C1 * m1 + C2 * m2).

Δt = 12.5 * 1320 * (1 – 0.2) / (920 * 0.05 + 4200 * 0.12) = 24 ºС.