Heat rate flow through bar? and thermal conductivity

omc1

1. Homework Statement A hot reservoir is maintained at a temperature of 78 oC, and a cold reservoir is maintained at a temperature of 5oC. They are separated by 11.00 cm, and are connected by a cylindrical bar made of two materials. The radius of the bar is 0.83 cm. The upper part of the bar is made of yellow brass (k = 220 W/(moC)), and has a length of 1.10 cm. The lower part's material is unknown. The temperature where the two materials are in thermal contact is 77.4875 oC. What is the rate of heat flow through the bar?
then find the thermal conductivity?

2. Homework Equations q=KA(deltaT)/L

3. The Attempt at a Solution i tried to use this equation to find the heat flow, and then I tried to set this equation to find the K for the other substance but its not working...thanks for help!!!

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Chestermiller

Mentor
1. Homework Statement A hot reservoir is maintained at a temperature of 78 oC, and a cold reservoir is maintained at a temperature of 5oC. They are separated by 11.00 cm, and are connected by a cylindrical bar made of two materials. The radius of the bar is 0.83 cm. The upper part of the bar is made of yellow brass (k = 220 W/(moC)), and has a length of 1.10 cm. The lower part's material is unknown. The temperature where the two materials are in thermal contact is 77.4875 oC. What is the rate of heat flow through the bar?
then find the thermal conductivity?

2. Homework Equations q=KA(deltaT)/L

3. The Attempt at a Solution i tried to use this equation to find the heat flow, and then I tried to set this equation to find the K for the other substance but its not working...thanks for help!!!
Not good enough. You need to show more of your work. I assume the hot reservoir is at the top, and the cold one is at the bottom. What are the temperatures at the two ends of the 1.1 cm yellow brass section? What is the rate of heat flow through this section?

omc1

yes the hot one is at the top, I tried it two different ways because I am confused about this problem but both ways i tried it got me similar answers, for the the heat flow I plugged in the numbers but for delta T I first used 77.4875, then I used 78-5, but both are wrong. for the second part I did KA(deltaT)/L= KA(deltaT)/L, but I sont understand this problem...

Chestermiller

Mentor
yes the hot one is at the top, I tried it two different ways because I am confused about this problem but both ways i tried it got me similar answers, for the the heat flow I plugged in the numbers but for delta T I first used 77.4875, then I used 78-5, but both are wrong. for the second part I did KA(deltaT)/L= KA(deltaT)/L, but I sont understand this problem...
I asked you what the temperatures are at the two ends of the 1.1 cm yellow brass section, and you haven't answered me. You will never get delta T correct if you are unable to answer this question.

This conduction problem is analogous to the problem of two electrical resistors connected in series across a voltage drop.

mpreds

Hi omc1,

Remember that you can view kA/L as a resistance, and the bar of differing materials would be "resistors" in series. So (k1A/L1) + (k2A/L2) = (kA/L)total

You will have to build three seperate equations from this information, combine and solve for k2 first then solve for your total q

-Mark

edit: Chestermiller beat me to it!

Chestermiller

Mentor
Hi omc1,

Remember that you can view kA/L as a resistance, and the bar of differing materials would be "resistors" in series. So (k1A/L1) + (k2A/L2) = (kA/L)total

You will have to build three seperate equations from this information, combine and solve for k2 first then solve for your total q

-Mark

edit: Chestermiller beat me to it!
Mark, you've kinda got it inverted. kA/L is a conductance, not a resistance. The equations you have written are for two bars in parallel, not series.

Chestermiller

Mentor
omc1:

To get the rate of heat flow through the bar (part A of your problem), you only need to focus on the 1.1 cm yellow brass portion. Can you solve the following problem:

I have a 1.1 cm long bar 0.83 cm in radius, made out of a material with a thermal conductivity of 220 W/(m C). The temperature at one end of the bar is held fixed at 78 C, and the temperature at the other end of the bar is held fixed at 77.4875 C. What is the rate of heat flow through the bar (in watts)?

omc1

sorry i did not see that question when you first posted it, the temperature at the top of the yellow brass is 78C and at the bottom of the yellow brass it is 77.4875C.
I used that for delta t and got the right answer...iam not sure why the cold temperature is not used because the bar is connected to the cold and the hot...
for the second part i set it up to solve for K2, i fount the Q for the yellow brass, but iam not sure about the total Q???

Chestermiller

Mentor
The exact same Q goes through the lower part of the bar. What are the temperatures at the two ends of the lower part of the bar? What is the Q through the lower part of the bar.

I hope you realize this problem is the same as taking two separate bars and gluing them together. You know what the temperatures are at the two ends of each of the two sections. Both sections have to have the same heat flow rate through them. After all, where else is the heat gonna go?

omc1

ok, Q through the lower part of the bar is the same as through the top par...the temperature at the lower end is 77.4875 and 5C. so Q=K2*A*deltaT/L2
L2=9.90cm, Q=2.22W, deltaT=77.4875-5

Chestermiller

Mentor
Yes. Now solve your equation for K2.

omc1

ok, double check me on this because it is saying that it is wrong, (2.218)*(.099)/(pi*.0083^2)*(77.4875-5)=13.997 W/mC

Chestermiller

Mentor
I get the same answer as you. I'm pretty confident we did it correctly.

omc1

yup we did, computer wanted more sig figs! thanks for your help!!

mpreds

Mark, you've kinda got it inverted. kA/L is a conductance, not a resistance. The equations you have written are for two bars in parallel, not series.
Thanks for the correction, my textbooks call it conduction resistance, which is why I called it resistance. I did have my resistance equations inverted for omc1 to integrate it into his/her equation more easily.

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