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?
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.
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