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 = m_{1} = 142 g =
0.142 kg

Mass of liquid in the second container = m_{2} = 25 g =
0.025 kg

Initial temperature of the liquid in the first container =
T_{1}

Initial temperature of the liquid in the second container =
T_{2} = 21 ^{o}C

Final temperature of the mixture = T_{3} = 42.6
^{o}C

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.

m_{1}C(T_{1} - T_{3}) =
m_{2}C(T_{3} - T_{2})

m_{1}(T_{1} - T_{3}) =
m_{2}(T_{3} - T_{2})

(0.142)(T_{1} - 42.6) = (0.025)(42.6 - 21)

T_{1} - 42.6 = 3.8

T_{1} = 46.4 ^{o}C

Initial temperature of the liquid in the first container = 46.4
^{o}C

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