Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.270 kg and a temperature of 61.0 ∘C.
How much ice at a temperature of -22.4 ∘C. must be dropped into the water so that the final temperature of the system will be 24.0 ∘C∘?

(Take the specific heat of liquid water to be 4190 J/kg⋅K, the specific heat of ice to be 2100 J/kg⋅K , and the heat of fusion for water to be 3.34×10⁵ J/kg)

Answer 1

Specific heat of water = C₁ = 4190 J/(kg.K)

Specific heat of ice = C₂ = 2100 J/(kg.K)

Latent heat of fusion of water = L = 3.34 x 10⁵ J/kg

Mass of the water = m₁ = 0.27 kg

Mass of the ice = m₂

Initial temperature of water = T₁ = 61 ^oC

Initial temperature of ice = T₂ = -22.4 ^oC

Melting point of ice = T₃ = 0 ^oC

Final temperature of the system = T₄ = 24 ^oC

Heat removed from the water to reach 24 ^oC= Q₁

Q₁ = m₁C₁(T₁ - T₄)

Heat added to the ice to reach 24 ^oC = Q₂

Q₂ = m₂C₂(T₃ - T₂) + m₂L + m₂C₁(T₄ - T₃)

Q₂ = m₂[C₂(T₃ - T₂) + L + C₁(T₄ - T₃)]

The heat removed from the water is equal to the heat added to the ice.

Q₁ = Q₂

m₁C₁(T₁ - T₄) = m₂[C₂(T₃ - T₂) + L + C₁(T₄ - T₃)]

(0.27)(4190)(61 - 24) = m₂[(2100)(0 - (-22.4)) + 3.34x10⁵ + (4190)(24 - 0)]

m₂ = 0.087 kg

Mass of ice to be dropped into the water = 0.087 kg