Homework Help: Rate of heat flow problem

1. May 11, 2010

huybinhs

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

A styrofoam cooler (k = 0.030 W/(m·°C) has outside dimensions of 0.190 m × 0.210 m × 0.340 m, and an average thickness of 2.2 cm. How long will it take for 3.60 kg of ice at 0°C to melt in the cooler if the outside temperature is 26.0°C?

2. Relevant equations

Delta Q / Delta t = [k A (T1-T2)] / l

3. The attempt at a solution

Delta Q = m L = 3.60 * 3.33*10^5 = 119880 cal.

=> Delta t = [Delta Q * l] / [k*A (T1 - T2)]

Delta t = 119880 *2.2*0.01 / 0.030*0.190*0.210*0.340*26 = 692.34 h which is wrong.

2. May 11, 2010

Mapes

3. May 11, 2010

huybinhs

Yes, I have checked the units but I have no idea where I am wrong!!!! ????

4. May 11, 2010

Mapes

cal W-1 m-1 is not units of time. It looks like the calorie part is just a typo, but think carefully about what area is used in the conduction equation.

5. May 11, 2010

huybinhs

Area = 0.190 m × 0.210 m × 0.340 m = 0.013566 m^3. Correct?

And what's up with the calc? :(

6. May 11, 2010

Mapes

Area isn't measured in cubic meters. What's the total cross-sectional area that the heat transfer occurs through?

And I don't get your calculation of "119880 cal."

7. May 11, 2010

huybinhs

Oh, Q = 1198800 cal

I have no ideas about the cross-sectional area ???

8. May 11, 2010

Mapes

This energy is still not correct, and I can tell that you really have not checked your constants and units for correctness. You're mixing up calories and Joules, and these are not equal.

If the heat were only being transferred through one side of the box, what would be the appropriate area? Now generalize this to the six-sided box.

9. May 11, 2010

huybinhs

Got ya, so:

Q = 1198800 J = 286520.076 calories.

and A = 6 *0.190 m × 0.210 m × 0.340 m = 0.081396 m^2.

Yes???

10. May 11, 2010

Mapes

Still looks like a volume to me (measured in cubic meters). Try finding the area of each side individually and adding them together.

11. May 11, 2010

huybinhs

Ok, let me try:

0.190 m × 0.210 m + 0.210 m × 0.340 m + 0.340m x 0.190m = 0.1759 m^2. Yes?

12. May 11, 2010

Mapes

Closer...

13. May 11, 2010

huybinhs

???? what do u mean closer?

Last edited: May 11, 2010
14. May 11, 2010

huybinhs

2 * 0.1759 = 0.3518 m^2 = area, correct?

15. May 11, 2010

huybinhs

Still stuck! Any one?

16. May 11, 2010

Mapes

Agreed.

17. May 11, 2010

huybinhs

Delta t = 1198800 *2.2*0.01 / 0.030*0.3518*26 = 96112.3 s = 1602 mins = 26.7 h which is wrong. How come ???

18. May 11, 2010

Mapes

Perhaps they want you to be a bit more precise with the area. Instead of using the area calculated from the outside of the container, try using the area calculated from the midpoint of the container walls. The answer is about 18% larger.

EDIT: Fixed typo.

Last edited: May 11, 2010
19. May 11, 2010

huybinhs

Sorry! I dont know how to calculate from the midpoint :(

I mean how?

20. May 11, 2010

huybinhs

You mean:

[0.190 m × 0.210 m + 0.210 × 0.340 m] / 2 = 0.05565 * 2 = 0.1113 , correct?