How many degrees does the temperature of an iron part weighing 8.4 kg

univerkov | February 21, 2021 | education

How many degrees does the temperature of an iron part weighing 8.4 kg rise if it is transferred the same amount of heat that is needed to heat 900gr of water by 15 °.

Data: m1 (weight of the iron part) = 8.4 kg; m2 (mass of water) = 900 g = 0.9 kg; Δt2 (change in water temperature) = 15 ºС.

Reference data: C1 (specific heat of iron) = 460 J / (kg * K); C2 (specific heat of water) = 4200 J / (kg * K).

Let us express the change in temperature of an iron part from the following equality: C1 * m1 * Δt1 = C2 * m2 * Δt2 and Δt1 = (C2 * m2 * Δt2) / (C1 * m1).

Calculation: Δt1 = (4200 * 0.9 * 15) / (460 * 8.4) = 14.67 ºС.

Answer: The temperature of the iron part will rise by 14.67 ºС.

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