What amount of heat was released during the cooling of water, the volume of which

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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