# 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.