A hot piece of copper is dropped into an ambient temperature oil bath. How do th

Question

A hot piece of copper is dropped into an ambient temperature oil bath. How do the entropy of the following systems change, increase, decrease or do not change? Copper Oil Copper + Oil Consider liquid water contained in a vessel as shown. The other part of the vessel is vacuumed and is several times larger than the filled compartment. The membrane between the two parts is ruptured with negligible energy. The vessel is insulated. Will the properties in the table increase (+), decrease (-) or not change (0)? Put your responses In the boxes below.

Explanation / Answer

a)

Copper will loose heat to the colder oil. Hence, its final temperature will be lower than the initial temperature.

ds = m*Cp*dT /T

Hence, dT will be negative. Hence, entropy will decrease.

Oil will gain heat from the hotter copper. Hence, its final temperature will be higher than the initial temperature.

ds = m*Cp*dT /T

Hence, dT will be positive. Hence, entropy will increase.

Copper + Oil entropy will increase since it is an isolated system.

b)

The entropy of water will increase as it expands even though the temperature will remain the same as there is no heat transfer.

ds = m*Cp*dT /T + RT dV /V

Here dT = 0, dV is positive due to expansion. Hence, ds = positive

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