# Homework Help: Heat problem (cooler at a beach)

1. May 15, 2015

### goonking

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

2. Relevant equations

Q = kAΔT/L

3. The attempt at a solution

so I need to find the volume of the cooler, and then the volume is the volume is ice in it?

then I need to find the surface area of 1 side of the cooler, than multiply it by 6. then plugging in the total area in Q = kAΔT/L

is this a good approach to the problem?

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2. May 15, 2015

### Raghav Gupta

Hey, that L is not latent heat in equation but length.

3. May 15, 2015

### goonking

oh, sorry! good catch

4. May 15, 2015

### Raghav Gupta

There is another equation for heat transfer what is that?

5. May 15, 2015

### goonking

Q = L m

we equate them so we get

L m = kAΔT/Length

we have L

we have k

we have A and L

I'm assuming ΔT is 29 C?

6. May 15, 2015

### Raghav Gupta

How to bring time term here which is 3 hrs?

7. May 15, 2015

### goonking

i think the answer we get, if we did all the algebra, is the amount of heat per seconds.

so we just convert 3 hours to seconds. then multiply by the heat per seconds

correct?

8. May 15, 2015

### Raghav Gupta

First of all the formula for heat conduction is
Q = kAtΔT/L where k is constant , A is area, t is time, ΔT is temperature difference and L is length.

9. May 15, 2015

### CWatters

In case it's not obvious... In this case L is the thickness of the insulation (eg the length of the heat path)

10. May 15, 2015

### goonking

Q = (0.035) (0.02m x 0.4m) (10800 seconds) (29 C) / (0.4m) = 219.24 joules

i believe we set Q = L m = 219.24 joules

m should be 6.56x10-4kg

correct?

11. May 15, 2015

### Raghav Gupta

Yes, it is correct.

12. May 15, 2015

### CWatters

No "A" is the surface area. You were right when you wrote..

L is the length of the heat path which is through the walls (eg 2cm).

13. May 15, 2015

### Raghav Gupta

Yeah, sorry.
The length to be taken is 2cm.

14. May 15, 2015

### goonking

oh, i forgot to multiply the area by 6, so it should be 6( 0.02m x 0.4m) = A

so new answer should be 26300 joules

i would have never thought the length was 2cm

15. May 15, 2015

### CWatters

kAΔT/L gives you the power (energy per second) flowing through the walls. If the walls are thicker (better insulated) then L is longer and less power flows. Make sense?

16. May 15, 2015

### goonking

ok, got the answer to be 1.57 kg, sounds much more reasonable now

17. May 15, 2015

### CWatters

Yes. I make the equation..

Q(Joules) = (0.035) (6 * 0.4m x 0.4m) (10800 seconds) (29 C) / (0.2m)

18. May 15, 2015

### goonking

should be 0.02 m

19. May 15, 2015

### Raghav Gupta

Yes you are right.

20. May 15, 2015

### CWatters

Yes sorry /0.02.

21. May 15, 2015

### goonking

make sure you put in 0.02 m, not 0.2m

22. May 15, 2015

### Raghav Gupta

Yes, sorry,
Edited.

23. May 15, 2015

### CWatters

Just a thought but..

The question isn't clear if 0.4m is the inside or outside dimension of the box.

24. May 15, 2015

### goonking

true, i was thinking that too when I tried to draw out the cooler

25. May 15, 2015

### Raghav Gupta

Isn't the answer given in your problem so you could verify?