Rate of heat flow problem

In summary: Because the sides of the current box are the midpoints of the walls of the previous box, the effective surface area is (0.190 m2 + 0.210 m2 + 0.340 m2)/(3.60 kg + 2.20 kg + 0.340 kg) = 0.1113 m2. So the answer is closer to 0.1146 m2.
  • #1
huybinhs
230
0

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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  • #2
Have you checked your units?
 
  • #3
Mapes said:
Have you checked your units?

Yes, I have checked the units but I have no idea where I am wrong! ?
 
  • #4
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.
 
  • #5
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? :(
 
  • #6
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."
 
  • #7
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 ?
 
  • #8
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.
 
  • #9
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.
 

1. What is the rate of heat flow?

The rate of heat flow refers to the amount of heat energy that is transferred per unit time. It is measured in units of joules per second, also known as watts.

2. How is the rate of heat flow calculated?

The rate of heat flow can be calculated using the formula Q/t, where Q is the amount of heat energy transferred and t is the time it takes for the heat transfer to occur. This formula is typically used for steady-state heat transfer problems.

3. What factors affect the rate of heat flow?

The rate of heat flow is affected by several factors, including the temperature difference between the two objects, the thermal conductivity of the materials involved, and the surface area and thickness of the objects.

4. How does the rate of heat flow impact temperature change?

The rate of heat flow plays a crucial role in determining the temperature change of an object. The higher the rate of heat flow, the faster the temperature of the object will change. This is because a higher rate of heat flow means more heat energy is being transferred in a shorter amount of time.

5. What are some real-life applications of rate of heat flow?

The rate of heat flow is an important concept in many fields, including thermodynamics, engineering, and meteorology. It is used to design heating and cooling systems, understand climate patterns, and study heat transfer in various materials and environments.

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