An aluminum calorimeter weighing 50 g is filled with 120 g of water and an electric heater with a power of 12.5

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%?

To find out the change in the temperature of an aluminum calorimeter with water, we use the formula: η = Apol / Azatr = (Sv * mv + Ca * mk) * Δt / (N * t), from where we can express: Δt = η * N * t / (Sv * mw + Ca * mk).

Const: St. – beats. heat capacity of water (Sv = 4200 J / (kg * K)); Ca – beats. heat capacity of aluminum (Ca = 920 J / (kg * K)).

Data: η – process efficiency (η = 80% = 0.8 (loss 20%)); N – heater power (N = 12.5 W); t is the duration of the heater operation (t = 22 min = 1320 s); mw is the mass of the poured water (mw = 120 g = 0.12 kg); mk is the mass of the calorimeter (mk = 50 g = 0.05 kg).

Calculation: Δt = η * N * t / (Sv * mw + Ca * mk) = 0.8 * 12.5 * 1320 / (4200 * 0.12 + 920 * 0.05) = 24 ºС.

Answer: The calorimeter with water should have heated up by 24 ºС.