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

1. Jun 21, 2010

### CharlieC89

3. 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

2. Jun 21, 2010

### 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. :)

3. Jun 25, 2010

### CharlieC89

thats great, thanks

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

4. Jun 25, 2010

### 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.