Question

A 0.035-kg ice cube at -30.0°C is placed in 0.315 kg of 35.0°C water in a very well-insulated container. The latent heat of fusion for water is Lf = 79.8 kcal/kg.

What is the final temperature of the water, in degrees Celsius?

Answer #1

Mass of the ice cube = m_{1} = 0.035 kg

Mass of the water = m_{2} = 0.315 kg

Specific heat of ice = C_{1} = 0.5
kcal/(kg.^{o}C)

Specific heat of water = C_{2} = 1
kcal/(kg.^{o}C)

Latent heat of fusion of water = L = 79.8 kcal/kg

Initial temperature of ice = T_{1} = -30
^{o}C

Initial temperature of water = T_{2} = 35
^{o}C

Melting point of ice = T_{3} = 0 ^{o}C

Final temperature of the water = T_{4}

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

m_{1}C_{1}(T_{3} - T_{1}) +
m_{1}L + m_{1}C_{2}(T_{4} -
T_{3}) = m_{2}C_{2}(T_{2} -
T_{4})

(0.035)(0.5)(0 - (-30)) + (0.035)(79.8) +
(0.035)(1)(T_{4} - 0) = (0.315)(1)(35 - T_{4})

0.525 + 2.793 + 0.035T_{4} = 11.025 -
0.315T_{4}

0.35T_{4} = 7.707

T_{4} = 22.02 ^{o}C

Final temperature of water = 22.02 ^{o}C

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