A 750 g copper calorimeter containing 200 g of water is in equilibrium at 20 deg

Question

A 750 g copper calorimeter containing 200 g of water is in equilibrium at 20 degree C. An experimenter now places 30 g of ice at 0 degree C in the calorimeter and encloses it with a heat insulating shield. The specific heat of copper is 0.418 J/g- degree C and that of water is 4.18 J/g- degree C. The heat of fusion of ice is 333 j/g. After the ice has melted and equilibrium has been attained, what will be the final temperature of the water? After the ice has melted and equilibrium has been attained, how much work must be supplied to the system (e.g. by means of a stirring rod) to restore all of the water to 20 degree C?

Explanation / Answer

Here,

1) let the final temperature is Tf

heat lost by caloriemeter + heat lost by water = heat gain by ice

750 * 0.418 * (20 - Tf) + 200 * 4.18 * (20 - Tf) = 30 * (333 + 4.18 * Tf )

solving for Tf

Tf = 10.2 degree C

the final equilibrium temperature is 10.2 degree C

part B)

work must be supplied = heat gain by system to raise the temperature to 20 degree C

work must be supplied = (750 * 0.418 + 230 * 4.18) * (20 - 10.2)

work must be supplied = 12494 J

the work must be supplied is 12494 J

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