To what temperature 15 liters of boiling water will cool if 2680 kJ is released into the atmosphere.

Given: V (volume of cooled boiling water) = 15 liters (in SI V = 15 * 10 ^ -3 m3); Q (heat released into the atmosphere) = 2680 kJ (2680 * 10 ^ 3 J).

Constants: Cw (heat capacity of boiling water, water) = 4200 J / (kg * K); ρ (density of boiling water) = 1000 kg / m3; tboil (boiling point of water) = 1000 ºС.

The temperature to which 15 liters of boiling water should cool down can be expressed from the formula: Q = Cw * m * (tboil – t) = Cw * ρ * V * (tboil – t), whence t = tboil – Q / (Cw * ρ * V).

Calculation: t = 100 – 2680 * 10 ^ 3 / (4200 * 1000 * 15 * 10 ^ -3) = 57.46 ºС.

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