How much heat is needed to melt a sample of lead weighing 0.237 kg at 26.4 C?
Assume the specific heat, the latent heat and the melting point of lead are 128 J/kg*C, 2.45*10^4 J/kg and 327.3 C respectively.
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A sample of lead weighing 0.237 kg is initially at 26.4 C. The heat required to melt the sample has to be determined. It is given that the melting point of lead is 327.3 C, the specific heat of lead is 128 J/kg*C, and the latent heat is 2.45*10^4 J/kg.
First the temperature of the sample has to be raised to 327.8 C. The heat required to do this is equal to 128*0.237*(327.3 - 26.4) = 128*0.237*300.9 = 9128.1 J
Once the sample is at the temperature at which lead changes from solid to heat additional heat has to be added to force the change in phase. This is equal to 2.45*10^4*0.237 = 5806.5 J
The total heat required to melt the lead sample is the sum of the heat required to raise the temperature and the heat required to cause the sample to melt. This is given as 9128.1 + 5806.5 = 14934.6 J
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