(adsbygoogle = window.adsbygoogle || []).push({}); 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* [tex]\Delta[/tex] T

Q_{fus}=m*c_{fus}

mass_{pb}=0.50 Kg @328^{o}C

c_{pb}=34 cal/Kg^{o}C

T_{final}=?

c_{fusion}=6800 cal

3. The attempt at a solution

Q_{fusion}=6800 cal/Kg * 0.50 Kg= 3400 cal

Q_{pb}= 3400 cal + .50Kg * 34 cal/Kg^{o}C* (328-x)

Q_{pb}=3400+5576-17x

For Water:

Q_{H2O}=2.5 Kg * 1000 cal/Kg^{o}C * (x-20^{o}C)

Q_{H2O}=2500x-50000

Setting these equal to each other I get:

2500x-50000=3400-17x+5576

2517x=58976 cal

x=23.4311^{o}C

The back of the book has the answer of 22^{o}C.

I can get 22^{o}C as an answer only if I subtract the latent heat of fusion:

2500x-50000=17x+5576

2517x=55576

x=22.0803^{o}C

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.

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# Homework Help: Specific Heat molten lead Question

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