What amount of heat was released during the cooling of water, the volume of which is 20 liters if the temperature changed from 100 to 50 * С?
V = 20 l = 20 * 10 ^ -3 m ^ 3. ρ = 1000 kg / m ^ 3. t1 = 100 “C. t2 = 50” C. C = 4200 J / kg * “C. Q -? When water is cooled, the amount of heat Q is released, which is determined by the formula: Q = C * m * (t1 – t2), where C is the specific heat capacity of water, m is the mass of water that cools down, t1, t2 are the initial and final water temperatures, respectively. The mass of water m is expressed by the formula: m = ρ * V, where ρ is the density of water, V is the volume of water. Q = С * ρ * V * (t1 – t2) Q = 4200 J / kg * “C * 1000 kg / m ^ 3 * 20 * 10 ^ -3 m ^ (100” C – 50 “C) = 4.2 * 10 ^ 6 J.
Answer: when the water was cooled, the amount of heat was released Q = 4.2 * 10 ^ 6 J.
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