Specific Heat molten lead Question

[Wellesley]

Homework Statement

If you pour 0.50 kg of molten lead at 328°C into 2.5 liters of water at 20°C, what will be the final temperatures of the water and the lead? The specific heat of (solid) lead has an average value of 3.4 X 10^-2 kcal/kg • °C over the relevant temperature range.

Homework Equations

Q=mass*c* \Delta T
Q_fus=m*c_fus
mass_pb=0.50 Kg @328^oC
c_pb=34 cal/Kg^oC
T_final=?
c_fusion=6800 cal

The Attempt at a Solution

Q_fusion=6800 cal/Kg * 0.50 Kg= 3400 cal
Q_pb = 3400 cal + .50Kg * 34 cal/Kg^oC* (328-x)
Q_pb=3400+5576-17x

For Water:
Q_H2O=2.5 Kg * 1000 cal/Kg^oC * (x-20^oC)
Q_H2O=2500x-50000

Setting these equal to each other I get:

2500x-50000=3400-17x+5576
2517x=58976 cal
x=23.4311^oC

The back of the book has the answer of 22^oC.

I can get 22^oC as an answer only if I subtract the latent heat of fusion:

2500x-50000=17x+5576
2517x=55576
x=22.0803^oC

The math seems pretty straight forward, so it makes me wonder whether the book made an error, or I did.

Any help would be appreciated! Thanks.

 

[ideasrule]

I think your answer is right. There's no reason to subtract the heat of fusion.

 

[Wellesley]

I think your answer is right. There's no reason to subtract the heat of fusion.

Thanks for the reply. This is the second time I've seen errors in the book, that's why I needed to make sure I hadn't made an error.

 

[Borek]

I would not say "subtract", I would say "ignore". But it has to be taken into account.

 

[meidew93]

I realize this is a year old. But I just did the same problem but my book says the specific heat of lead is 140 J/kg*°C. Just in case somebody else reads this.