Why Does a Copper Block's Internal Energy Remain Stable Despite Continuous Work?

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The discussion centers on the phenomenon of a copper block's internal energy remaining stable despite continuous work being applied at 85W. It is established that as the temperature of the block increases, the rate of heat loss to the surroundings also increases, leading to a point where heat loss equals heat input, resulting in temperature stabilization. Factors such as reduced friction between the block and the rubbing pad and potential thermal equilibrium are considered, but the primary conclusion is that heat loss is proportional to the temperature difference between the block and its surroundings.

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there's a copper block being rubbed with a pad and constant work is being done on it at 85W..starts at room temperature and t=0

i) from t=250 to t=300 almost no internal energy is gained by the block though work is being done on it. Explain this.

there's a graph given (temp against time) a curve starts (0,0) levels off at (80,250)

i wrote..As temperature increases the difference between the block and surroundings increase so the rate at which heat is lost from block increase.
(exponentially) I am not sure..

or prehaps the frictional force between the two (block and pad reduced) like when using sandpaper on wood goes smooth?

could some kinda thermal equilibrium been reached between the block and pad...same internal energy? or all/most internal energy of pad transferred to block is that possible?

(Q=U + W) work is being done on the block so W is negitive and internal energy is gained so its positive..would there be a point where they equal each other in magnitudes so they cancel leaving Q as '0'...i don't know..help
non of my answers seem right for some reason.

(uggh I am really annoyed i had to write this out 3 times now because it won't send decided best to copy it this time just in case)
 
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Heat loss is proportional to [itex]\displaystyle{ \Delta T}[/itex] so as the temperature of the block increases it loses more heat to the surroundings. At some point the heat loss must equal the heat input. When that point is reached the temperature will stabilize.
 
thank you.
 

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