Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.350 kg and a temperature of 78.3 ∘C. How much ice at a temperature of -24.2 ∘C must be dropped into the water so that the final temperature of the system will be 40.0 ∘C ?

Answer 1

Step-by-step explanation:

Given:

c₁ = 2100 J / (kg·°C)

c₂ = 4200 J / (kg·°C)

λ₁ = 3.4·10⁵ J / kg -

t₁ = -24.2°C

t₀ = 0°C

t₂ = 78.3°C

t = 40.0°C

m₂ = 0.350 kg

___________

m₁ - ?

Ice is heated to 0°C:

Q₁ = c₁·m₁·(t₀ - t₁) = 2100·m₁·(0 - (-24.2)) = 50 820·m₁ J

The ice is melting:

Q₂ = λ·m₁ = 340 000·m₁ J

Cold water is heated:

Q₃ = c₂·m₁·(t - t₀) = 4200·(40.0 - 0)·m₁ = 168 000 J

Hot water is cooling down:

Q₄ = c₂·m₂·(t₂ - t) = 4200·0,350·(78.3 - 40.0) = 56 300 J

Thermal badance equation:

Q₁ + Q₂ + Q₃ = Q₄

m₁·(50820 + 340 000 + 168 000) = 56 300

m₁·(3 620 000) = 56 300

Mass of ice:

m₁ = 56 300 / 3 620 000 ≈ 0,016 kg or 16 g