heat transfer
i have a few problems with the question.
1. "thermal conductivity is high " means majority of the heat is conducted and very low amount is absorbed and therfore emitted so how can we find out change of temp of A .
2. temp between the surfaces of A and B is to be taken 0K ??
3. masses are not given so how can we find out rate of change of temp
10 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.I = σe∆(T⁴) should work considering it's blackbody radiation, and I guess emissivity is one because prefect blackbodies.
That gives you rate of change of energy, so dQ = msdT after that
Sorry for the really late reply
nahi bana
can u solve the numberwise queries
Thermal conductivity is high probably just means that heat distribution on the surface is uniform at all times
Temp between the surface is irrelevant because radiation doesn't care about a medium
And you don't need mass because heat capacity is given, not specific heat capacity
in pg its mentioned "it is given that the thermal conductivity of material A is very high so we can assume that it will not absorb any heat thus the rate of heat loss by A is given as dq/dt= A sigma T^4
oh right , but then how will we determine temp of surrounding
Why do we need that?
newtons law of cooling lagane ke lie ?
nvm i was thinking smthing else ig
all g?
will close this soon