How much heat will be released when cooled by 80 degrees C of a lead part weighing 400 g?
| August 5, 2021 | education
The amount of heat released when cooling the lead part:
Q = C * m * Δt, where C is the specific heat of lead (C = 140 J / (kg * K)), m is the mass of the lead part (m = 400 g = 0.4 kg), Δt is the temperature change of the lead part (Δt = 80 ºС).
Q = C * m * Δt = 140 * 0.4 * 80 = 4480 J.
Answer: When the lead part is cooled, 4480 J of heat will be released.
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