Heat transfer, single and double paned window. Stove and pot.

[Ulyaoth]

A glass window 1m x 2m and .005m thick has thermal conductivity of 1.4W/m K. Inner surface temp of 15C and outside of -20C, what is the heat loss through the glass.

double paned construction two panes separated by .01m, if glass surfaces are 10C and -15C, with air thermal conductivity of .024W/m K
q''= -k dT/L
q’’= 1.4W/m•K((288K-253K)/(.005m)), q’’=9800W/m2, q=9800W/m2 • 2m2 = 19600W

Double Paned: Rtot = L/kg+L/ka+L/kg,
Rtot = (.005m)/(1.4W/(m•K))+(.010m)/(.024W/(m•K))+(.005m)/(1.4W/m•K)
Rtot=0.424K/W,
q=ΔT/Rtot, q=(283K-253K)/(.0424K/W), q=70.75W


It just seemed like too big of a difference to be the correct answer. Just one more I wasn't sure of.

The 5mm thick bottom of a 200mm diameter pan may be copper(k = 390W/m K) or aluminum (240W/m K). When used to boil water the surface of the bottom exposed to the water is nominally 110C. IF the heat transferred from the stove to the pot is 600W, what is the temperature of the surface in contact with the stove for the two materials
A=0.031416m2, q''=19100W
q=k(Tbottom-Tin)/L (q•L)/k+Tin=Tbottom,alum
Tbottom,alum=(19100W/m^2(.005m))/(240W/m•K)+383K = 383.398K
Tbottom,copper=(19100W/m^2(.005m))/(390W/m•K)+383K = 383.245K


Again, just wasn't sure, it seemed the numbers were too close.

 

[LawrenceC]

q=ΔT/Rtot, q=(283K-253K)/(.0424K/W), q=70.75W

You have a slight error in the above. It should be 82.6. The .0424 should be .424.

The huge difference comes from the additional resistance of the air. Look at the ratios of k/dx for air gap and for a pane of glass. The difference is large.

In a real life problem, you also have the radiation factor where the hotter pane loses heat to the cooler pane. It has an effect but is not considered in this problem. To complicate the issue further, there is motion of the trapped gas between the panes that further enhances the heat transfer.

 

[LawrenceC]

Second part dealing with the pot on the stove is correct.

 

[Ulyaoth]

Thank you, just wanted to make sure I had the right idea for the problem.

 

[DeeNos]

I would like to provide a response to the above content by first acknowledging that heat transfer is a complex phenomenon and there are many factors that can affect the accuracy of our calculations. However, based on the given information and assuming ideal conditions, here are my thoughts on the two scenarios:

1. Heat loss through a single-paned glass window:
Using the formula q''= -k dT/L, we can calculate the heat loss through the glass window to be 19,600W. This seems like a reasonable result, considering the significant temperature difference between the inner and outer surfaces of the window and the relatively high thermal conductivity of glass.

2. Temperature of the bottom surface of a pot:
For the second scenario, we are given the heat transfer rate of 600W and the surface temperature of 110C for the bottom of the pot. By using the formula q=k(Tbottom-Tin)/L, we can solve for the temperature of the surface in contact with the stove for both copper and aluminum materials. The results of 383.398K for aluminum and 383.245K for copper seem reasonable, as the difference between the two is only 0.153K. However, it is important to note that these calculations assume ideal conditions and there may be other factors that could affect the accuracy of our results, such as heat loss to the surroundings.

In conclusion, while the numbers may seem close, they are within a reasonable range and reflect the basic principles of heat transfer. As a scientist, it is important to consider all the variables and limitations when making calculations and interpreting results.