Heat capacity

1. Jun 19, 2014

HemaZ

1. The problem statement, all variables and given/known data

the answer is c) but i can't figure why !! is the more heat capacity substance lose and gain energy faster than the other one so it take more energy to get it's temperature raised by the same amount because it lose most of it very fast during the heating process ?

2. Jun 19, 2014

CWatters

It takes energy to melt wax. How much energy does each ball contain?

3. Jun 19, 2014

HemaZ

i think Metal B will contain the double of Metal A

4. Jun 19, 2014

Orodruin

Staff Emeritus
And therefore ...

5. Jun 19, 2014

BvU

So B can melt more wax than A.

I had fun imagining both balls being "placed above the surface of a wax block" at considerable height and on e.g. a shelf. For the sake of seriousness I suppose they mean "placed on the surface of a wax block"

Also liked the realistic touch added by mentioning the 20 cm separation...

6. Jun 20, 2014

CWatters

I assume you mean double the energy rather than double the metal.

7. Jun 20, 2014

HemaZ

yes sir

8. Jun 20, 2014

adjacent

Do not think. Use the specific heat capacity equation and confirm it.
$E=mc\Delta t$ -$t$ is temperature.

9. Jun 20, 2014

HemaZ

i know that the metal B contain double energy of metal A but the energy transfer in the same time period to the wax is the double too ?!!

10. Jun 20, 2014

adjacent

Of course. Why not?
If we assume the temperature difference between the metal and the wax is same. Then that's it.

11. Jun 20, 2014

HemaZ

does this mean the substance with higher heat capcity transfer heat faster ?

12. Jun 20, 2014

Orodruin

Staff Emeritus
This would depend both on the specific heat capacity and the thermal conductivity of the substance. The specific heat tells you how much energy you need to put into (or take away from) the material to change its temperature. The thermal conductivity tells you how large the energy flow is relative to the temperature gradient.

Given the same thermal conductivity and transfer conditions on the surface, the material with the higher specific heat capacity would cool down slower, meaning it would initially melt wax at more or less the same rate. However, it would stay warm for longer and therefore keep on melting wax and go deeper in the end. For different thermal conductivities, the time-scale of the process may be different, but the total energy release from the material to end up below the wax melting point will stay the same.

13. Jun 20, 2014

BvU

Why bother about faster if the exercise asks for a distance ? They sink in and they stop at some point. Don't have to stop at the same moment !

14. Jun 20, 2014

Orodruin

Staff Emeritus
Why stop at what the exercise asks you for if you find other aspects of the problem interesting or wonder about related things? I would recommend to rather encourage such things.

15. Jun 20, 2014

BvU

@druin: there may be more exercises waiting. Hopefully with more concrete input. Thermal conductivity of most metals is a lot higher than that of wax, so a temperature gradiënt in the metal balls is pretty small anyway.

16. Jun 23, 2014

Vibhor

Really nice explaination .

Do you mean that both A and B will have to reach the temperature of wax i.e change in temperature would be same (Δt is equal) and since specific heat of B is more than that of A ,using Q=msΔt,heat transfer will be more for B ?

17. Jun 23, 2014

BvU

@Vib: more or less. The initial 400 degrees is pretty hefty.
I don't plan to do the experiment, but I can fantasize a little bit:
first choose my balls, e.g.

Indium, cp 0.24 kJ/(kg.K) , rho 7310 kg/m3 and
Steel with cp 0.49 kJ/(kg.K), rho 7850 kg/m3 ), then my target:
Paraffin, cp some 2.5 kJ/(kg.K), rho 900 kg/m3
this wax melts at, say 55 °C, heat of fusion a hefty 200 kJ/kg !

The 20 cm apart from the OP comes in handy for our shopping list:
Let's play safe and buy a lump of wax 60 x 40 x 15 cm3. Just about luggable.

Metal balls ? say 5 cm diameter. If we copy adjacent's formula mcΔt, cooling them from 400 to 55 degrees gives us some 40 and 87 kJ of energy in the form of heat (a ratio of 1:2 if you squint a bit).

For the paraffin, heating up a one-ball volume from 20 to 55 degrees costs only 5.2 kJ, but then also melting it another 11.8 kJ ! together 17 kJ.

The Indium ball can heat up and melt about 2.3 ball volumes and the steel ball 5.1 of them.
So (cylinder / sphere and the last bit is a half sphere) a depth of 1.7 and 3.5 ball diameters is my best guess. Ultimately, after the heat is evenly distributed, the 127 kJ heats up the whole block (mcΔt again) only 1.6 degrees, so there is no risk of a room-size paraffin splotch.

So far, I would bet on the steel ball coming through and the Indium getting stuck a little below half-way.

One thing I would ask your input on: behind the ball, the wax solidifies again, thus releasing 2/3 of the heat going from solid 20 to liquid 55 degrees. A nice fraction of that heat is passed on to the ball, so it may well go a significant factor deeper than I calculated. Who dares to bet the Indium ball makes it through ?

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