Need help with a question involving heat engines, the carnot cycle and entropy.

[CharlieC89]

The Attempt at a Solution

For
a) I imagine it is easier to control the temperature of the hot reservoir because you could use a constant heat source e.g. bunsen burner.

c) melting/freezing water.

d) Entropy of the environment increases and the entropy of the universe always increases.

Any help with b) or e) would be massively appreciated

 

[dingo_d]

b)

\eta=\frac{W}{Q_{in}}=1-\frac{|Q_{out}|}{Q_{in}}=1-\frac{T_{min}}{T_{max}}

If you raise the temp of hot reservoir by \Delta T you'll have: T_{max}'=T_{max}+\Delta T, and if you lower the cold reservoir by the same amount you'll have:
T_{min}'=T_{min}-\Delta T.
So the new efficiencies will be:
\eta_1=1-\frac{T_{min}}{T_{max}'}=\frac{T_{max}-T_{min}+\Delta T}{T_{max}+\Delta T}
and
\eta_2=1-\frac{T_{min}'}{T_{max}}=\frac{T_{max}-T_{min}+\Delta T}{T_{max}}
If you rearrange those two:
\eta_1(T_{max}+\Delta T)=\eta_2T_{max}

So it is obvious that \eta_1>\eta_2 the efficiency of Carnot cycle is greater when you take away the temp from cold reservoir. :)

 

[CharlieC89]

thats great, thanks

anybody have any idea for e), even just a prod in the right direction would be excellent.

 

[hikaru1221]

For e), the condition for minimum work to be achieved is the coefficient of performance \eta = \frac{dQ_{cool}}{dA} = \frac{T_{cool}}{T_{hot}-T_{cool}}
Where:
dA is the work needed to lower the temperature by dT
dQ_{cool} is the heat taken from the water/ice.
T_{hot} is the temperature of the sink, which is the air in this problem.
T_{cool} is the temperature of the source, which is the water/ice. Notice that T_{cool} varies during the process.

There is another equation relating dQ_{cool} with dT (actually 2 equations, each corresponds to one process). Compute the integral, and you will obtain the total work needed.