Energy, 370.5 J is used to heat a sample of lead from 257 to 327 degrees celcius where it promptly melts and stays at 327 degree. What is the mass?   So my teacher said use the equation for heating and melting.. so q=mL AND q=mCdeltaT .. BUT i dont get how to do that.

Expert Answers

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You need some more data as well. The heat capacity of lead. Cp and the Latent heat of melting of lead.

Latent heat of melting of lead = L = 22.4 kJ/kg

Specific heat capacity of lead Cp= 130 J/(kgK)

Now the melting point of lead is 327 Celcius. So to melt it we have to heat from 257 C to 327 C while it is solid and from there we have to supply the latent heat of melting to lead, so it will become liquid.

So the total energy supplied is 370.5 J. Let th mass of lead is m



Total energy supplied = Total energy supplied to heat from 257 to 327 + Latent heat of melting supplied

Q = mCpdt + mL

370.5 J = m x 0.13 J/gK x (327-257) K + m x 22.4 J/g

370.5 = 9.1 m + 22.4 m

m = 11.762 g.


The mass of lead is 11.762 g.




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