Question

You have two containers of the same liquid. The first container has 142.0 g at T1°C and the second has 25 g at 21°C. In order to consolidate and save space, you mix the two liquids into one container and find that the two portions have now reached an equilibrium temperature of 42.6°C. What was the initial temperature of the liquid in the first container? °C

Answer 1

Mass of liquid in the first container = m₁ = 142 g = 0.142 kg

Mass of liquid in the second container = m₂ = 25 g = 0.025 kg

Initial temperature of the liquid in the first container = T₁

Initial temperature of the liquid in the second container = T₂ = 21 ^oC

Final temperature of the mixture = T₃ = 42.6 ^oC

Specific heat of the liquid = C

The heat lost by the liquid in the first container is equal to the heat gained by the liquid in the second container.

m₁C(T₁ - T₃) = m₂C(T₃ - T₂)

m₁(T₁ - T₃) = m₂(T₃ - T₂)

(0.142)(T₁ - 42.6) = (0.025)(42.6 - 21)

T₁ - 42.6 = 3.8

T₁ = 46.4 ^oC

Initial temperature of the liquid in the first container = 46.4 ^oC