What amount of heat is released during the transformation of 250 g of water, having a temperature

What amount of heat is released during the transformation of 250 g of water, having a temperature of 40 degrees C, into ice at 0 degrees C?

To find how much heat is released when water of mass m turns into ice, it is necessary to use the formula: Q = Q₀ + Q₁, where Q₀ is the amount of heat released when water is cooled from the initial temperature t₀ to the melting (crystallization) temperature t, Q₁ is the amount of heat released during its crystallization. Since Q₀ = с ∙ m ∙ | t – t₀ |, where с is the specific heat capacity of water, and Q₁ = λ ∙ m, where λ is the specific heat of melting of ice, then:

Q = m ∙ (s ∙ | t – t₀ | + λ).

It is known that the specific heat capacity of water is c = 4200 J / (kg ∙ ºС), the specific heat of melting of ice λ = 340 kJ / kg = 340000 J / kg, and the melting point of ice is t = 0 ºС. We took m = 250 g = 0.25 kg of water having a temperature of t₀ = 40 ° C. We get:

Q = 0.25 kg ∙ (4200 J / (kg ∙ ºС) ∙ 40ºС + 340,000 J / kg);

Q = 127000 J = 127 kJ.

Answer: when water turns into ice, an amount of heat equal to 127 kJ will be released.