Question

# A 0.035-kg ice cube at -30.0°C is placed in 0.315 kg of 35.0°C water in a...

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?

Mass of the ice cube = m1 = 0.035 kg

Mass of the water = m2 = 0.315 kg

Specific heat of ice = C1 = 0.5 kcal/(kg.oC)

Specific heat of water = C2 = 1 kcal/(kg.oC)

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

Initial temperature of ice = T1 = -30 oC

Initial temperature of water = T2 = 35 oC

Melting point of ice = T3 = 0 oC

Final temperature of the water = T4

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

m1C1(T3 - T1) + m1L + m1C2(T4 - T3) = m2C2(T2 - T4)

(0.035)(0.5)(0 - (-30)) + (0.035)(79.8) + (0.035)(1)(T4 - 0) = (0.315)(1)(35 - T4)

0.525 + 2.793 + 0.035T4 = 11.025 - 0.315T4

0.35T4 = 7.707

T4 = 22.02 oC

Final temperature of water = 22.02 oC

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