How to Calculate Heat Loss in a Freezer Door Handle

In summary, you need to find the power lost per hour through two bolts, in watts, and then multiply it by 3600 to get the total power loss over the course of an hour.
  • #1
Hendon100
3
0

Homework Statement


This is my first post on these forums, this problem is part of a lab and its part my end of year result but we never did anything on it in class so i have'nt a clue how to do it. Help would be really appreciated

The handle of a freezer door 12cm thick is attached by two brass bolts 0.75cm in diameter that pass through the entire door and are secured on the inside by nuts. The interior of the freezer is maintained at -32 degrees celsius and the room temperature is 34 degrees celsius; the thermal conductivity is 109W.M-1.K-1. Find the heat lost per hour through the bolts.



Homework Equations



Like i said, i haven't a clue. I don't know why we even were given this question.

The Attempt at a Solution

 
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  • #2
Welcome to PF!

Hi Hendon100! Welcome to PF! :smile:

You don't need to know anything about heat loss …

just look at the units! :wink:

you're told the power loss of the material in watts per metre per ºK, and power times time = energy, so (remembering to use both the diameter and the length) the energy loss in this case is … ? :smile:
 
  • #3
Thanks for the help, I used this equation. I don't know if I am right though.

Power
= (thermal conductivity)(C.S.A)(difference in temp)
=
Distance​

Thermal conductivity= 109
C.S.A= (3.75x10-3)2(3.14)= 4.42x10-5m2
Difference in temp=34-(-32)=66
Distance= .12m

Power=
(109)(4.42x10-5)(66)
= 2.65 Watts per bolt
.12​

I then multiplied this by 2 because there's 2 bolts. So 5.3 Watts overall

And then i just used the equation Power= Energy x Time

5.3(3600)= 19080 joules lost per hour

:uhh::uhh: which seems a bit high for two little brass bolts but i could be wrong. Please reply to tell me if i did it right, and if not how do i do it. Thanks alot.
 
  • #4
(you meant Energy = Power x Time :wink:)

Yes, it does look high, but the maths seems fine. :smile:
 
  • #5
alright, thank you very much:smile::smile:
 

1. What is thermal conductivity?

Thermal conductivity is a measure of how well a material can conduct heat. It is the property that determines how quickly heat can pass through a material.

2. How is thermal conductivity measured?

Thermal conductivity is typically measured using a device called a thermal conductivity meter. This device measures the amount of heat that flows through a material over a given time and distance. The resulting value is known as the thermal conductivity coefficient.

3. What factors affect thermal conductivity?

There are several factors that can affect thermal conductivity, including the type of material, its density, moisture content, and temperature. Generally, materials with higher density and lower moisture content tend to have higher thermal conductivity. Additionally, thermal conductivity can also vary with temperature, with some materials exhibiting higher conductivity at higher temperatures.

4. Why is thermal conductivity important in scientific research?

Thermal conductivity is an important property in scientific research because it helps us understand how heat flows through different materials. This information can be used to design more efficient heating and cooling systems, as well as to study the thermal properties of various materials.

5. What is a common thermal conductivity problem encountered in scientific research?

One common thermal conductivity problem is the determination of thermal conductivity in complex or composite materials. These materials may have varying densities and compositions, making it difficult to accurately measure thermal conductivity. Additionally, thermal conductivity can also be affected by external factors such as pressure and magnetic fields, which can add further complexity to the problem.

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