# Homework Help: Specific Heat molten lead Question

1. Jan 24, 2010

### Wellesley

1. The problem statement, all variables and given/known data

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.

2. Relevant equations
Q=mass*c* $$\Delta$$ T
Qfus=m*cfus
masspb=0.50 Kg @328oC
cpb=34 cal/KgoC
Tfinal=?
cfusion=6800 cal
3. The attempt at a solution

Qfusion=6800 cal/Kg * 0.50 Kg= 3400 cal
Qpb = 3400 cal + .50Kg * 34 cal/KgoC* (328-x)
Qpb=3400+5576-17x

For Water:
QH2O=2.5 Kg * 1000 cal/KgoC * (x-20oC)
QH2O=2500x-50000

Setting these equal to each other I get:

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

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

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

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

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.

2. Jan 24, 2010

### ideasrule

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

3. Jan 24, 2010

### Wellesley

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.

4. Jan 25, 2010

### Staff: Mentor

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

5. Dec 12, 2011

### 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.