Question

A piece of lead weighs 0.2kg. It takes 300 seconds for it to reach melting point of 600 degrees C from an initial temperature of 0 degrees C. The specific heat capacity of lead is 130 J kg^-1 k^-1. Calculate the total energy supplied to the lead before it starts to melt.

Answer 1

melting point of lead is 600 K and not °C

Answer 2

\[\Delta H=\int\limits_{T1}^{T2} cp dT\]

Answer 3

DH= 130 Jkg-1K-1 * (600-0) = 78000 Jkg-1
DH= 78 kJkg-1 * 0,2kg = 15,6 kJ

Answer 4

and
q= DH/t= 15,6 kJ/300s = 52 W

Answer 5

Is DH the same as U = m c delta T?

Answer 6

well yes, i dont know what letter you use to mark it...

Answer 7

Ok. For the first part of the question. Q is the energy supplied or removed. M = mass in kg, C = specific heat capacity and Delta T is the change in temperature.

Answer 8

Could you explain it that way. I do not understand your terminology.

Answer 9

what do you need explained?

Answer 10

DH= 130 Jkg-1K-1 * (600-0) = 78000 Jkg-1. Is this the answer to the first part of the question. I have done this calculation. U = m c delta T. 0.2*130Jkg-1 K-1 *600K. Is that correct?

Answer 11

yes it is correct and if you compare it i did the same

Answer 12

Ok great. Now what about the second part of question when i ask for the rate of heating of the lead?

Answer 13

well taht is q in W and W is J/s

Answer 14

heat flow rate

Answer 15

I have'nt done any work on heat flow rate to be honest!

Answer 16

well then it's time to start! :)

Answer 17

(1) Q = kA delta T / d ?

Answer 18

well that is different equation...

Answer 19

What is the formula for heat flow rate ?

Answer 20

dont you have your textbooks or scripts for studying?

Answer 21

Yes i have advanced physics for you. But i cannot find any formula for the heat flow rate in it.

Answer 22

ok lets think about that for a second, how do you mark heat? lets say Q and how do you mark time, lets say t and rate of heat transfer is Q/t and lets say we mark that with h, or q and there you have it...

Answer 23

i must admit that its been about 5 years i studied that, nowdays i just get DSC or DMA or FTIR or some other method diagrams and dont have to deal with that...

Answer 24

Ok. Q is energy supplied or removed / T = change in temperature (rise and fall) ?

Answer 25

you take heat and take its change in time and you get amount of heat given to the body or heat that body releases in time...

Answer 26

it is like speed generaly, so if "s" is distance and "t" is time so s/t is speed of travel...

Answer 27

So it will be Q / T = ?. Is that it

Answer 28

yes but time is usually marked with small t

Answer 29

Ok. So it will be Q/ delta t = 15600J / 300s = 52J ?

Answer 30

yes

Answer 31

and it is J/s or J s-1 or W

Answer 32

Ok thanks it looks more simple than it should be!