Question

You have 200 g of coffee at 45 ⁰C, Coffee has the same specific heat as water. How much ice at -10 ⁰C do you need to add in order to reduce the coffee’s temperature to 30 ⁰

Answer 1

Specific heat of water = Specific heat of coffee = C₁ = 4186 J/(kg.^oC)

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

Latent heat of fusion for water = L = 334000 J/kg

Mass of the coffee = m₁ = 200 g = 0.2 kg

Mass of ice added = m₂

Initial temperature of the coffee = T₁ = 45 ^oC

Initial temperature of the ice = T₂ = -10 ^oC

Melting point of ice = T₃ = 0 ^oC

Final equilibrium temperature = T₄ = 30 ^oC

The heat lost by the coffee is equal to the heat gained by the ice.

m₁C₁(T₁ - T₄) = m₂C₂(T₃ - T₂) + m₂L + m₂C₁(T₄ - T₃)

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

(0.2)(4186)(45 - 30) = m₂[(2100)(0 - (-10)) + 334000 + (4186)(30 - 0)]

m₂ = 0.02613 kg

m₂ = 26.13 g

Mass of ice to be added to the coffee = 26.13 g