During the crystallization of molten tin, taken at the melting temperature, and its subsequent cooling to 32 ºС

During the crystallization of molten tin, taken at the melting temperature, and its subsequent cooling to 32 ºС, an amount of heat of 315 kJ was released. What is the mass of tin?

Initial data: t (final cooling temperature) = 32 ºС; Q (heat released during crystallization) = 315 kJ = 315 * 10 ^ 3 J.

Reference data: λ (specific heat of fusion of tin) = 59 * 10 ^ 3 J; C (specific heat of tin) = 230 J / (kg * K); tmelt (tin melting point) ≈ 232 ºС.

The released amount of heat: Q = λ * m + C * m * (tpl – t) = m * (λ + C * (tpl – t)), whence m = Q / (λ + C * (tpl – t)) …

Let’s calculate: m = 315 * 10 ^ 3 / (59 * 10 ^ 3 + 230 * (232 – 32)) = 3 kg.