How Much Lead Can Be Melted with 1000 Joules Starting at 20°C?

[mawalker]

This is the question...

What is the maximum mass of lead you could melt with 1000 J of heat, starting from 20C?

I know that for lead
Tm = 328 degrees C
Lf (j/k) = 0.25 x 10 ^5
Tb = 1750 degrees C
Lv = 8.58 x 10^5

but I'm not really sure what i need to use. Q = Mc delta T

but how can you figure this without having an ending temperature? I'm confused. Anyone have an idea?

 

[Doc Al]

Hint: Assume the lead has just melted. What's its final temperature?

Hint2: Think of there being two steps:

(1) Raising the temperature
(2) Melting the lead​

 

[mawalker]

the lead begins to melt at 328 degrees celsius. so this would be the final temperature?

 

[Doc Al]

Sounds right to me.

 

[mawalker]

ok so delta T = 308. c of lead = 128, setting up my equation 1000J = M(128)(308)... 1000 = 39424M M = .02536 kg. M = 25.36 g and this is the wrong answer.

 

[Doc Al]

ok so delta T = 308. c of lead = 128, setting up my equation 1000J = M(128)(308)... 1000 = 39424M M = .02536 kg. M = 25.36 g and this is the wrong answer.

Looks like you only considered the first step (raising the temperature) and not the second (the actual melting). The total energy of those two steps equals 1000 J.

 

[mawalker]

i don't really understand. what would be the change in temperature during the melting? wouldn't it just be zero? or would it just keep increasing?

 

[Doc Al]

During the state change (step 2) the temperature doesn't change, but energy is still required to melt a given mass of lead. Consider the latent heat of fusion.