A 26.4 g silver ring (cp = 234 J/kg·°C) is heated to a temperature of 66.2°C and then placed in a calorimeter containing 4.94 ✕ 10-2 kg of water at 24.0°C. The calorimeter is not perfectly insulated, however, and 0.136 J of energy is transferred to the surroundings before a final temperature is reached. What is the final temperature?

The final temperature of the mixture = 64.834 °C.

Explanation:

Heat lost by the silver ring = heat gained by the water + heat transferred to the surrounding.

c₁m₁ (t₁-t₃) = c₂m₂ (t₃-t₂) + Q ... Equation 1

Where c₁ = specific heat capacity of the silver copper, m₁ = mass of the silver copper, t₁ = initial temperature of the silver copper, t₃ = final temperature of the mixture. c₂ = specific heat capacity of water, t₂ = initial temperature of water, m₂ = mass of water, Q = energy transferred to the surrounding.

making t₃ the subject of the equation,

t₃ = [c₁m₁t₁+c₂m₂t₂-Q] / (c₁m₁+c₂m₂) ... Equation 2

Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92*10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

t₃ = [ (234*26.4*66.2) + (4200*0.0492*24) - 0.136]/[ (234*26.4) + (4200*0.0492) ]

t₃ = (408957.12+4959.36-0.136) / (6177.6+206.64)

t₃ = (413916.48-0.136) / 6384.24

t₃ = 413916.34/6384.24

Thus the final temperature of the mixture = 64.834 °C.