Determinant of how much heat is required to heat a mixture of 300 g of water and 50 g of alcohol from 20 * to 70 * C.

Problem data: m1 (mass of water in the mixture) = 300 g (in SI m1 = 0.3 kg); m2 (mass of alcohol) = 50 g (in SI m2 = 0.05 kg); t0 (initial temperature of the created mixture) = 20 ºС; t (final temperature) = 70 ºС.

Constants: C1 (heat capacity of water) = 4200 J / (kg * K); C2 (heat capacity of alcohol) = 2500 J / (kg * K).

We calculate the heat for heating the created mixture by the formula: Q = (C1 * m1 + C2 * m2) * (t – t0).

Calculation: Q = (4200 * 0.3 + 2500 * 0.05) * (70 – 20) = 69 250 J.

Answer: 69 250 J of heat is required.