How much heat must be spent in order to obtain water at 100 ° C from 5 kg ice taken at a temperature (-10 ° C)?

How much heat must be spent in order to obtain water at 100 ° C from 5 kg ice taken at a temperature (-10 ° C)? Specific heat of ice 2100 J / (kg * ° C), its melting point 0 ° C, specific heat of melting of ice 340 kJ / kg, specific heat of water 4200 J / (kg * ° C), boiling point of water 100 ° C Specific heat of vaporization water 2.3 MJ / kg.

To heat ice with a mass of m from the initial temperature t₀ to the melting temperature t количества, you need the amount of heat Q₀ = c₀ ∙ m ∙ (t₁ – t₁), where c₀ is the specific heat capacity of ice. To melt the ice, you need the amount of heat Q₁ = λ ∙ m, where λ is the specific heat of melting of ice. To further heat the water to the boiling point t₂, you will need the amount of heat Q₂ = с ∙ m ∙ (t₂ – t₁). In total you will need:

Q = Q₀ + Q₁ + Q₂ or Q = m ∙ (с₀ ∙ (t₁ – t₀) + λ + с ∙ (t₂ – t₁)).

It is known from the problem statement that ice with a mass of m = 5 kg was heated from a temperature of t₀ = – 10 ° C and water was obtained at a boiling point of t₂ = 100 ° C. Specific heat of ice c₀ = 2100 J / (kg ∙ ° C), its melting point t₁ = 0 ° C, specific heat of melting of ice λ = 340 kJ / kg = 340000 J / kg, specific heat of water c = 4200 J / (kg ∙ ° C). Then:

Q = 5 kg ∙ (2100 J / (kg ∙ ° C) ∙ (0 ° C – (- 10 ° C)) + 340,000 J / kg + 4200 J / (kg ∙ ° C) ∙ (100 ° C. – 0 ° C));

Q = 3905000 J ≈ 3.9 MJ.

Answer: it is necessary to spend ≈ 3.9 MJ.