How much heat will be released during the crystallization of 200 g of aluminum and cooling it to room temperature?

univerkov | May 31, 2021 | education

Initial data: m (mass of aluminum) = 200 g; tк (room temperature) = 20 ºС.

Reference data: tmelt (melting point of aluminum) = 660 ºС; λ (specific heat of fusion) = 3.9 * 10 ^ 5 J / kg; C (specific heat) = 920 J / (kg * K).

SI system: m = 200 g = 200/1000 kg = 0.2 kg.

Let’s compose a formula for calculating the released heat: Q = Q1 + Q2 = λ * m + C * m * (tpl – tк).

Let’s perform the calculation: Q = 0.2 * 3.9 * 10 ^ 5 + 920 * 0.2 * (660 – 20) = 195 760 J.

Answer: 195,760 Joules of heat will be released.

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