Question

You have a 2.00 kg block of lead. Lead melts at 327.5°C. CPb = 130.0 J/kg x g C and Hf for lead is 2.04 × 10^4 J/kg. You begin at room temperature (25.0°C). How much heat do you need to transfer to melt all the lead?

Answer 1

@roadjester

Answer 2

What do all of your letters mean? I think Hf is latent heat of fusion....

Answer 3

@ybarrap

Answer 4

yes hf means heat of fusion

Answer 5

I'm just learning thermodynamics, but let's see if we can work this out.

Is \(C_{Pb}\) the heat capacity for Lead?

We probably need to convert C to K for temp.

What do you think are the relevant equations?

Is it \(U=nC_{Pb}\Delta T\) Joules?

Answer 6

*If we only care about differences, we don't need to convert C to K.

Answer 7

yeah my teacher sees every little detail so she wants every conversion and units to be perfect

Answer 8

@roadjester

Answer 9

I haven't done thermodynamics in a while since it was two semesters ago i think

Answer 10

You're going to need two equations for this.
Q=mc\(\Delta\)T
and
Q=mL
Where the first one is when you are changing temperature and the second is changing between states.

Answer 11

Ok, what the heck is CPb?

Answer 12

I'm guessing Pb is lead but what is CPb?

Answer 13

i think the Cpb means the heat capacity for lead like @ybarrap said

Answer 14

m=2kg
\(\Large c_{Pb}=130...\) \(\color {red} {what \space\space the \space\space heck\space\space is\space\space up\space\space with\space\space the\space\space units?}\)
\(\large L_f(Latent \space heat\space of\space \space fusion)\)=2.04 X10^4 J/kg
T_i =25 degrees C

Answer 15

you've got Joules per, gram, kilogram, AND C??? four units?

Answer 16

yep i think so I've got : 2.00 kg block of lead
lead melts at 327.5 C
Cpb = 130.0 J/kg x g degrees C
Hf = 2.04 x 10^4 J/kg for lead

Answer 17

there is no way you can have two units of mass

Answer 18

start at room temperature of 25 degrees celsius
then find how much heat is needed to transfer to melt the lead

Answer 19

I know that, I've taken thermophysics

Answer 20

but your units are screwy

Answer 21

yeah it looks weird maybe they messed up on it

Answer 22

i think the g needs to be taken out and just remain with kg

Answer 23

either g or kg, can't have both

Answer 24

$$
\large{
U_2-U_1=m\cdot c \cdot (T_2-T_1)\\
=2~kg\times 130.0~ J/kg\cdot K \times(327.5-25) \\
=78,650~J\\
(Q_{lead})_{in}=h_f \times m_{Pb}=2.04 × 10^4~J/kg\times2kg=40,800 J\\
}
$$
http://hyperphysics.phy-astr.gsu.edu/hbase/tables/sphtt.html

The change in internal energy in changing the temperature of the lead from 25 to 327.5 C is 78,650 J. However, only 40,800 J is required to melt all the lead. The remaining energy will be stored in the liquid lead.

Answer 25

I may be wrong here.

It might be that 78,650 J is required to melt the lead because you need to get to the temp of 327.5 C to melt. Once melting is complete, 40,800 J would have been released through latent heat of fusion and the remaining energy stored in the liquid.

Answer 26

yes they ask me for the amount of heat needed to be transferred to melt the entire lead

Answer 27

but i think what you got was correct

Answer 28

I think my 1st answer is correct, 40,800 J. I just checked here and get approximately the same answer -- I also validated for gold -

http://wiki.answers.com/Q/How_much_energy_is_it_required_to_melt_2_kg_of_gold#slide=2

Answer 29

for where u put kg is it a specfici amount or just kg?

Answer 30

I used this latent heat of fusion chart for other elements - http://www.engineeringtoolbox.com/fusion-heat-metals-d_1266.html

BTW, your heat of fusion for lead is 2.04 × 10^4 J/kg while the site above shows it to be 2.3 × 10^4 J/kg, which really doesn't matter that much. But when I used the 2.3 × 10^4 J/kg, I got the exact same answer as wiki.answers.com for lead, 36.8 kJ, versus my answer of 40.8 kJ.

Answer 31

I don't understand your question

Answer 32

does kg stand for a specific number or is it just supposed to be written there?

Answer 33

I just wrote the units (for example, kg) so you are aware what the number preceding it is supposed to represent.

Answer 34

just helps with dimension analysis, to validate that units cancel appropriately

Answer 35

The only problem I have with this solution, if it is in fact 40.8 kJ is that this did not require us to use temperature at all. Perhaps if we found that there was not enough energy generated by the change in internal temperature, then we could not say that all the lead was melted. So in a sense, we DID use that information.

Answer 36

so was 78,650 supposed to be part of the answer?

Answer 37

No.

I think that if THAT number, taking into account temperature change (i.e. the change in internal energy) were less than 40.8 kJ, then we could not say that ALL of the lead would have melted.

Answer 38

It IS part of the answer to prove that the change in internal energy was higher than 40.8 kJ, which was required to melt ALL the lead.

Answer 39

so 78,650 is the heat required to be transferred to melt all the lead?

Answer 40

No, the heat required to melt ALL the lead is just 40.8kJ.

The other number was the ACTUAL heat that was transferred. So the lead received more heat than was required to melt it.

Answer 41

Remember, you can heat a solid until it melts. And then you can add more heat after it is melted. This additional heat will just be stored in the liquid until there is enough heat to turn the liquid into a gas.

That's what was done here -- but we did not check if it started turning into a gas.

Answer 42

oh ok yeah it just asks for the amount of heat needed to be transferred

Answer 43

so my answer would be 78,650?

Answer 44

No, the heat required to melt ALL the lead is just 40.8kJ.

Answer 45

so 40,800 kJ is final answer to "How much heat must you transfer to melt all the lead?"

Answer 46

40,800 J because kg's cancel out* sorry

Answer 47

Yes.