Determine how much heat will be released during the crystallization of 3.1 liters of mercury.

Determine how much heat will be released during the crystallization of 3.1 liters of mercury. The specific heat of crystallization of mercury is 0.12⋅105 J / kg, the density of mercury is 13600 kgm3.

Given: V = 3.1L, [lambda] = 0.12 * 10 ^ 5 J / kg, p = 13600 kg / m ^ 3
SI: V = 0.0031 m ^ 3
Find: Q
Decision:
1. Q = [lambda] m
m = pV
m = 13600 kg / m ^ 3 * 0.0031 m ^ 3 = 42.16 kg
Q = 0.12 * 10 ^ 5 J / kg * 42.16 kg = 505920J
Answer: Q = 505920J