To cool 2 kg of water from 30C to 12C, pieces of ice are thrown into the water at 0C.

To cool 2 kg of water from 30C to 12C, pieces of ice are thrown into the water at 0C. how much ice is required to cool the water?

Let ice pieces of mass m₂ with specific heat of fusion λ = 340000 J / kg at t₂ = 0 ° С.

At the same time, warm water gives off an amount of heat to the ice:

Q₁ = m₁ ∙ s ∙ (t₁ – t).

Ice during melting and subsequent heating of melt water to t = 12 ° C takes away the amount of heat from warm water:

Q₂ = λ · m₂ + m₂ ∙ s ∙ (t – t₂).

To determine how much ice is required to cool the water, we compose the heat balance equation:

Q₁ = Q₂ or m₁ ∙ s ∙ (t₁ – t) = λ m₂ + m₂ ∙ s ∙ (t – t₂).

We get that:

m₂ = m₁ ∙ s ∙ (t₁ – t) / (λ + s ∙ (t – t₂)) or

m₂ = (2 kg ∙ 4200 J / (kg ∙ ° С) ∙ (30 ° С – 12 ° С)): (340,000 J / kg + 4200 J / (kg ∙ ° С) ∙ (12 ° С – 0 ° WITH));

m₂ = 0.387 kg.

Answer: You need 0.387 kg of ice.