Question

A 21.0 g silver ring (cp = 234 J/kg·°C) is heated to a temperature of 81.6°C and then placed in a calorimeter containing 5.20 ✕ 10-2 kg of water at 24.0°C. The calorimeter is not perfectly insulated, however, and 0.140 J of energy is transferred to the surroundings before a final temperature is reached. What is the final temperature? °C

Answer 1

Mass of the silver ring = m₁ = 21 g = 0.021 kg

Mass of the water in the calorimeter = m₂ = 5.20 x 10^-2 kg = 0.052 kg

Initial temperature of the silver ring = T₁ = 81.6 ^oC

Initial temperature of the water = T₂ = 24 ^oC

Final equilibrium temperature = T₃

Specific heat of silver = C₁ = 234 J/(kg.^oC)

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

Heat lost to the surroundings = Q_L = 0.14 J

Heat lost by the silver ring = Q₁

Q₁ = m₁C₁(T₁ - T₃)

Heat gained by the water = Q₂

Q₂ = m₂C₂(T₃ - T₂)

The heat lost by the silver ring is equal to the heat gained by the water plus the heat lost to the surroundings.

Q₁ = Q₂ + Q_L

m₁C₁(T₁ - T₃) = m₂C₂(T₃ - T₂) + Q_L

(0.021)(234)(81.6 - T₃) = (0.052)(4186)(T₃ - 24) + 0.14

400.98 - 4.914T₃ = 217.672T₃ - 5224.128 + 0.14

222.586T₃ = 5624.968

T₃ = 25.27 ^oC

Final temperature = 25.27 ^oC