Question

A 25.0-g block of ice at -15.00°C is dropped into a calorimeter (of negligible heat capacity) containing water at 15.00°C. When equilibrium is reached, the final temperature is 8.00°C. How much water did the calorimeter contain initially? The specific heat of ice is 2090 J/kg ∙ K, that of water is 4186 J/kg ∙ K, and the latent heat of fusion of water is 33.5 × 104 J/kg.

Answer 1

Specific heat of ice = C₁ = 2090 J/(kg.K)

Specific heat of water = C₂ = 4186 J/(kg.K)

Latent heat of fusion of water = L = 33.5 x 10⁴ J/kg

Mas of the block of ice = m₁ = 25 g = 0.025 kg

Mass of water in the calorimeter = m₂

Initial temperature of the ice = T₁ = -15 ^oC

Initial temperature of the water = T₂ = 15 ^oC

Final equilibrium temperature = T₃ = 8 ^oC

Melting point of ice = T₄ = 0 ^oC

The heat energy gained by the ice is equal to the heat energy lost by the water.

m₁C₁(T₄ - T₁) + m₁L + m₁C₂(T₃ - T₄) = m₂C₂(T₂ - T₃)

(0.025)(2090)(0 - (-15)) + (0.025)(33.5x10⁴) + (0.025)(4186)(8 - 0) = m₂(4186)(15 - 8)

m₂ = 0.341 kg

Mass of water in the calorimeter initially = 0.341 kg