# Specific Heat Capacity-

1. May 24, 2010

### pat666

1. The problem statement, all variables and given/known data
3.00 kg of molten lead is allowed to cool until is has solidified. It is found that the temperature of the lead falls from 605 K to 600 K in 10 s, remains constant at 600K for 300 s, and then falls to 595 K in a further 8.4 s. Assuming that the loss of heat energy remains constant and the specific heat capacity of solid lead is 140 Jkg-1 K-1, calculate:

(a) the rate of loss of energy from the lead; ()
(b) the specific latent heat of fusion of lead; ()
(c) the specific heat capacity of liquid lead.

2. Relevant equations

3. The attempt at a solution
No idea how to start this, can someone give me a hand to start please.

2. May 24, 2010

### irycio

Now, what you know is that Q/t=const. Lets mark it with a letter a, but it doesn't really matter. Hence Q/t=a => Q=a*t.
You are also given specific heat capacity of solid lead. You have the additional data on solid lead in thae last part of your sentence, where it cools down from 600K to 595. Now, mass during cooling als remains constant. Now, since you know how fast the solid lead cooled down from 600K to 595K and you can calculate the amount of heat Q''' neaded for this, you can calculate our 'a', since Q/t remains constant. Now, having this, you can calculate everything else.

3. May 24, 2010

### pat666

So Q=140*10? Thats not what your saying i doubt but im not sure on what a is exactly?
Q=mc*deltaT i have m and delta T and c so i found Q to be 2100 for the first cooling but this is not constant is it????

4. May 25, 2010

You calculated the heat lost by the solid lead whilst it cooled.The rate of heat loss will be your value(2100J) divided by the time(8.4s).This tells you how much heat is lost in one second and the question tells you to assume that this rate has remained constant.

5. May 25, 2010

### pat666

arrrrr ok thanks that makes sense

6. May 26, 2010

### pat666

hey for the answers did yous get a)-210J/s ,B)-2100J/kg and c) 117.6J/(kg*K)

7. May 28, 2010

### EzaMoo

I get a) 2100 / 8.4 = 250J/s
b) Lf = Q/m = 2100 / 3 = 700 J/kg
c) I am unsure of tho.... would it be c = Q/m deltaT = 700 / 3 x 278 = 0.84 J/kgK

8. May 28, 2010

### pat666

a) 2100/10
you used all the times in the wrong sections i think

9. May 28, 2010

### EzaMoo

I was unsure of which instance of cooling to use but the two posters previously mentioned the second instance of cooling which took 8.4s.

Also, how did you get your answer for c? Did you do anything similar to me?

10. May 28, 2010

### pat666

2100J is Q for the temperature drop from 605K to 600K which took 10s. for that you used the SHC of solid lead ---- so t has to be 10s. making your part b wrong and your part c wrong.

dont know what you did for c, but what i did was
Q/t=-210j/s and the time it took to cool was 8.4s (thats where 8.4 is used) so Q=-210*8.4s from there q=mcdelta T calculate c......

you agree with that???,,,,,,, im not absolutely sure on this but it seems logical.

11. May 28, 2010

### pat666

ok i see where i went wrong i found the wrong Q(i used the SHC of solid lead when it was liquid././ stupid), now i think your a) is right but my procedure for c) is right i think..

12. May 28, 2010

Hello folks.I think you might see this problem more clearly if you sketch a graph of how the temperature varies with time.Your graph has three sections:
1.In the first ten seconds there is liquid lead only and its temperature is dropping.
2.At ten seconds the liquid starts to solidify and the temperature levels off.It takes three hundred seconds for all of the liquid to change to solid.
3.When all of the liquid has solidified the temperature stars to drop again

Look again at the third section of the graph.The heat lost =3*140*5=2100J.
This heat is lost in 8.4s so the heat lost in 1 second=2100/8.4=250W(J/s)
Knowing how much heat is lost in one second you can calculate how much heat is lost by the liquid lead in the first ten seconds and by the lead whilst it is changing to a solid in the next three hundred seconds.You can take it from there.

13. May 28, 2010

### XxDonniexX

ur answers are all different to mine
i got a) -250J/s b) -25000J/kg c) 166.67J/kg/K
Its Tony btw

14. May 28, 2010

### pat666

tony your a is definitely right im redoing b and c. so ill tell you if i agree once i have done them

15. May 28, 2010

### EzaMoo

Hey Tony,

After all the advise, I re-did section b and c.

Bingo.... I get the same :)

Thanks for all your help guys!

16. May 28, 2010

### pat666

Yea i got the same - once i fixed my mistake.

17. May 28, 2010

### XxDonniexX

hey pat wat did u get for the US submarine's velocity in question 2b.

18. May 28, 2010

### pat666

fricken huge 65m/s but thats right!!

19. May 29, 2010

### XxDonniexX

lol i got 68.52 m/s

20. May 29, 2010

### pat666

yeah i rounded that right