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

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SUMMARY

The discussion focuses on calculating the time required for 3.60 kg of ice at 0°C to melt in a Styrofoam cooler with specific dimensions and thermal conductivity (k = 0.030 W/(m·°C)). The initial calculation resulted in an incorrect time of 692.34 hours due to errors in unit conversions and area calculations. Participants emphasized the importance of using the correct cross-sectional area for heat transfer and suggested calculating the effective surface area at the midpoint of the cooler's walls for more accurate results. The final corrected time calculation yielded approximately 26.7 hours, highlighting the need for precision in thermal calculations.

PREREQUISITES
  • Understanding of heat transfer principles, specifically conduction.
  • Familiarity with the thermal conductivity of materials (e.g., Styrofoam).
  • Knowledge of unit conversions between calories and Joules.
  • Ability to calculate surface areas of geometric shapes, particularly boxes.
NEXT STEPS
  • Learn about effective thermal conductivity and its application in heat transfer calculations.
  • Study the differences between specific heat capacities of ice and water.
  • Explore the concept of thermal resistance in multi-layer materials.
  • Investigate the use of numerical methods for solving heat transfer problems in complex geometries.
USEFUL FOR

Students in thermodynamics, engineers working on thermal management solutions, and anyone involved in designing insulated containers or studying phase change materials.

huybinhs
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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!
 
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Have you checked your units?
 
Mapes said:
Have you checked your units?

Yes, I have checked the units but I have no idea where I am wrong! ?
 
huybinhs said:
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.
 
Mapes said:
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? :(
 
huybinhs said:
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."
 
Mapes said:
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 ?
 
huybinhs said:
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.
 
Mapes said:
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?
 
  • #10
Still looks like a volume to me (measured in cubic meters). Try finding the area of each side individually and adding them together.
 
  • #11
Mapes said:
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?
 
  • #12
Closer...
 
  • #13
Mapes said:
Closer...

? what do u mean closer?
 
Last edited:
  • #14
Mapes said:
Closer...

2 * 0.1759 = 0.3518 m^2 = area, correct?
 
  • #15
Still stuck! Any one?
 
  • #16
huybinhs said:
2 * 0.1759 = 0.3518 m^2 = area, correct?

Agreed.
 
  • #17
Mapes said:
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 ?
 
  • #18
huybinhs said:
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.
 
Last edited:
  • #19
Mapes said:
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?
 
  • #20
Mapes said:
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?
 
  • #21
Now it's all wrong! I'm really confused! ?
 
  • #22
Should I use Q = m c DeltaT which c in ice = 2100 J/kg*C or c in liquid = 4186 ?
 
  • #23
Anyone?
 
  • #24
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.
 

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