A 3.50 kg block of copper at a temperature of 87 °C is dropped into a bucket con

Question

A 3.50 kg block of copper at a temperature of 87 °C is dropped into a bucket containing a mixture of ice and water whose total mass is 1.30 kg. When thermal equilibrium is reached the temperature of the water is 4 °C. How much ice was in the bucket before the copper block was placed in it?(ci = 2000 J/(kg.°C), cw = 4186 J/(kg.°C), Lf=3.35 × 105 J/kg, Lv=2.26 × 106 J/kg, ccopper = 387 J/(kg.°C). Neglect the heat capacity of the bucket.) A 3.50 kg block of copper at a temperature of 87 °C is dropped into a bucket containing a mixture of ice and water whose total mass is 1.30 kg. When thermal equilibrium is reached the temperature of the water is 4 °C. How much ice was in the bucket before the copper block was placed in it?(ci = 2000 J/(kg.°C), cw = 4186 J/(kg.°C), Lf=3.35 × 105 J/kg, Lv=2.26 × 106 J/kg, ccopper = 387 J/(kg.°C). Neglect the heat capacity of the bucket.) A 3.50 kg block of copper at a temperature of 87 °C is dropped into a bucket containing a mixture of ice and water whose total mass is 1.30 kg. When thermal equilibrium is reached the temperature of the water is 4 °C. How much ice was in the bucket before the copper block was placed in it?(ci = 2000 J/(kg.°C), cw = 4186 J/(kg.°C), Lf=3.35 × 105 J/kg, Lv=2.26 × 106 J/kg, ccopper = 387 J/(kg.°C). Neglect the heat capacity of the bucket.)

Explanation / Answer

The specific heat of Cu is = 387 J/kg•K;
a temperature drop of 83 K means an energy loss of
387 J/kg•K * 3.5 kg * 83 K = 112423.5 J = 26.86 kcal.

The energy needed to heat 1.3 kg of water from 0°C to 4°C is
(1.3 kg)(1 kcal/kg•°C)(4°C) = 5.2 kcal, so subtract this from the 26.86 kcal initially available, meaning approximately 21.66 kcal were used to melt the ice.

The energy needed to melt ice at 0°C is 79 kcal/kg;
21.66 kcal/(79 kcal/kg) = 0.274 kg.

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