How Long Does It Take for Ice to Melt in a Styrofoam Cooler?

[huybinhs]

Homework Statement

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?

Homework Equations

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

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.

Please advise! Thanks!

 

[Mapes]

Have you checked your units?

 

[huybinhs]

Have you checked your units?

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

 

[Mapes]

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

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.

 

[huybinhs]

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.

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

And what's up with the calc? :(

 

[Mapes]

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

And what's up with the calc? :(

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."

 

[huybinhs]

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."

Oh, Q = 1198800 cal

I have no ideas about the cross-sectional area ?

 

[Mapes]

Oh, Q = 1198800 cal

I have no ideas about the cross-sectional area ?

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.

 

[huybinhs]

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.

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?

 

[Mapes]

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

 

[huybinhs]

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

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?

 

[Mapes]

Closer...

 

[huybinhs]

Closer...

? what do u mean closer?

 

[huybinhs]

Closer...

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

 

[huybinhs]

Still stuck! Any one?

 

[Mapes]

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

Agreed.

 

[huybinhs]

Agreed.

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 ?

 

[Mapes]

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 ?

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.

 

[huybinhs]

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 10% larger.

Sorry! I don't know how to calculate from the midpoint :(

I mean how?

 

[huybinhs]

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 10% larger.

You mean:

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

 

[huybinhs]

Now it's all wrong! I'm really confused! ?

 

[huybinhs]

Should I use Q = m c DeltaT which c in ice = 2100 J/kg*C or c in liquid = 4186 ?

 

[huybinhs]

Anyone?

 

[Mapes]

Picture a box like the one in the problem. One way to calculate its surface area for heat transfer is to sum up the areas of the outer sides, as you did. Another way is to sum up the areas of the inner sides (i.e., subtract all the wall thicknesses from the side lengths). But perhaps a more precise way is to sum up the effective surface area for a box with sides at the midpoints of the walls of the current box. It's a way of splitting the difference between the inner and outer surface area values.