What amount of heat is released during the crystallization of 3kg of tin taken at the crystallization

What amount of heat is released during the crystallization of 3kg of tin taken at the crystallization temperature due to then cooling it to a temperature of 25 ° C

Tin, mass m, being in a liquid state at the melting temperature t₁, solidifying and cooling down to the temperature t₂, will release the amount of heat Q = Q₁ + Q₂. During solidification, the amount of heat will be released Q₁ = λ ∙ m, where λ is the specific heat of fusion; upon cooling, Q₂ = c ∙ m ∙ (t₂ – t₁) will be released, where c is the specific heat capacity. Then:

Q = λ ∙ m + с ∙ m ∙ (t₂ – t₁).

For tin λ = 59000 J / kg, c = 230 J / (kg ∙ ° C). It is known from the problem statement that we took tin with a mass of m = 3 kg at a crystallization temperature t₁ = 232 ° С and then cooled it to a temperature t₂ = 25 ° С. Substitute the values ​​of the quantities into the calculation formula:

Q = 59000 J / kg ∙ 3 kg + 230 J / (kg ∙ ° С) ∙ 3 kg ∙ | 25 ° С – 232 ° С |;

Q ≈ 320,000 J = 320 kJ.

Answer: 320 kJ will be allocated.