• Sudikshya Pant
In summary: I will show my work below:In summary, a solid cylindrical copper rod is inserted into a large block of solid hydrogen at its melting temperature, while its other end is exposed to thermal radiation from surrounding walls at a higher temperature. The sides of the rod are insulated and there is no energy loss or gain except at the ends of the rod. Using the equations H = Aes(T^4-T_s^4) and H = dQ/dt = kA(T^4-Tc^4)/L, the equilibrium temperature of the blackened end is found to be 14.26 K. This is calculated by approximating (500^4-T^4) as 500^4, and using a value of 1
Sudikshya Pant

## Homework Statement

One end of a solid cylindrical copper rod 0.200 m long and 0.0250 m in radius is inserted into a large block of solid hydrogen at its melting temperature, 13.84 K. The other end is blackened and exposed to thermal radiation from surrounding walls at 500.0 K. (Some telescopes in space employ a similar setup. A solid refrigerant keeps the telescope very cold—required for proper operation— even though it is exposed to direct sunlight.) The sides of the rod are insulated, so no energy is lost or gained except at the ends of the rod. (a) When equilibrium is reached, what is the temperature of the blackened end? The thermal conductivity of copper at temperatures
near 20 K is 1670 W/m^2 K. b) At what rate (in kg>h) does the solid hydrogen melt?

## Homework Equations

H = Aes(T^4-T_s^4)H =dQ/dt =kA(T^4-Tc^4)/L

## The Attempt at a Solution

I tried to solve it by taking the equilibrium temperature as T and equated the two equations above thinking that the rate at which the radiation is absorbed by the black end equals to the rate at which the heat is conducted from that end to the colder end. But that gives me wrong answer. I can't seen to understand the equilibrium and thus, have to admit don't understand what to do and why I should do it. Thus, please help me solve as well as understand this problem. Thank you.

Last edited by a moderator:
Have another look at your heat conduction equation. Does it involve T4 terms?

CWatters
mjc123 said:
Have another look at your heat conduction equation. Does it involve T4 terms?
No. I solved it without using the power four. Sorry, for the error here. (I can't even edit it now, can I?). Actually, I even kept the answer of T in the above equations but they don't seem to give equal values. Thus, I came to a conclusion that my method is wrong. The answer is 14.26K.

Your approach seems to be correct - I get 14.26 that way - but as you haven't shown your working it's not possible to identify where the mistake is.
Simplifying hint: As T << 500, approximate (5004 - T4) by 5004.

I would like to know what value did you use for 'e' of copper. Am I wrong to use 0.3 for it?

Wait...that's where I went wrong. I should use 1 for "black" end of copper right?

I used 1 and got the right answer.

## 1. What is the difference between radiation and conduction?

Radiation is the transfer of heat through electromagnetic waves, such as from the sun. Conduction is the transfer of heat through direct contact between objects.

## 2. How does steady state heat flow occur?

In steady state heat flow, the rate of heat transfer remains constant over time, meaning the amount of heat entering a system is equal to the amount of heat leaving the system.

## 3. What factors affect the rate of heat transfer through radiation?

The rate of heat transfer through radiation is affected by the temperature difference between two objects, the surface area of the objects, and the emissivity of the objects (how well they emit and absorb radiation).

## 4. How is heat flow through conduction calculated?

The rate of heat transfer through conduction is calculated using Fourier's law, which states that the rate of heat transfer is proportional to the temperature difference, the area of contact, and the thermal conductivity of the materials involved.

## 5. Can heat transfer occur through both radiation and conduction at the same time?

Yes, in most situations heat transfer occurs through a combination of radiation and conduction. For example, a cup of hot coffee will lose heat through radiation to the surrounding air, but also through conduction to the table it is sitting on.

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