What amount of heat does ice absorb during melting, if its initial temperature is -1 s?

To determine how much heat is absorbed by ice with mass m during melting, it is necessary to take into account that the ice must first be heated from the initial temperature t₀ to the melting temperature t₁. To do this, you need the amount of heat Q₀ = c ∙ m ∙ (t₁ – t с), where c is the specific heat capacity of ice. Further, in order to melt the ice, you need the amount of heat Q∙ = λ ∙ m, where λ is the specific heat of melting of ice. In total you need:

Q = Q₀ + Q₁ or Q = m ∙ (s ∙ (t₁ – t₀) + λ).

From the reference tables we find that the specific heat of ice c = 2100 J / (kg ∙ ° C), its melting point t₁ = 0 ° C, the specific heat of melting of ice λ = 340 kJ / kg = 340,000 J / kg. It is known that if its initial temperature was t₀ = – 1 ° С. We get:

Q = m ∙ (2100 J / (kg ∙ ° C) ∙ (0 ° C – (- 1 ° C)) + 340,000 J / kg);

Q = m ∙ 342100 J.

Answer: each kilogram of this ice, when melted, will absorb an amount of heat equal to 342.1 kJ

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