How much will the temperature of water with a volume of 100 liters change if 80% of the heat released during

How much will the temperature of water with a volume of 100 liters change if 80% of the heat released during the combustion of charcoal weighing 0.625 kg is used to heat it?

Given: V (volume of water) = 100 l (in SI V = 0.1 m3); η (percentage of heat transfer) = 80% (0.8); m2 (mass of charcoal) = 0.625 kg.

Constants: ρw (density of heated water) = 1000 kg / m3; according to the condition Sv (specific heat capacity of water) = 4.2 kJ / (kg * K) = 4.2 * 10 ^ 3 J / (kg * K); q (beats heat of combustion of charcoal) = 3.4 * 10 ^ 7 J / kg.

We express the change in water temperature from the equality: Sv * ρw * V * Δt = η * q * m ^ 2; Δt = η * q * m ^ 2 / (Sv * ρw * V) = 0.8 * 3.4 * 10 ^ 7 * 0.625 / (4.2 * 10 ^ 3 * 1000 * 0.1) ≈ 40, 5 ° C.