A problem that uses the result we derived for the efficiency of the Carnot cycle as a function of $T_{H}$ and $T_{L}$. Don't forget to pay attention to the correct units for temperature.

Another Carnot cycle problem, but this time you will have to look at the derivation rather than just the end results.

Start from conservation of energy

$W=Q_{H}-Q_{L}$

and use the equations for the two heats to arrive at an expression that links the work done to the difference in temperature.

Now you will need to find a way to change this equation so that you only have one temperature and all other terms are known.

You can find an expression that links the ratio of the temperatures $T_{H}$ and $T_{H}$ to the ratio of the volumes $V_{C}$ and $V_{D}$ by considering that

$P_{B}V_{B}=nRT_{H}$ $P_{C}V_{C}=nRT_{L}$ $P_{B}V_{B}^{\gamma}=P_{C}V_{C}^{\gamma}$

The last equation gives

$\frac{P_{B}V_{B}}{P_{C}V_{C}}=\frac{V_{C}^{\gamma-1}}{V_{B}^{\gamma-1}}=(\frac{V_{C}}{V_{B}})^{\gamma-1}$

which can be used with the first two to get an expression only in terms of temperatures and volumes.

This time we are looking at the Otto cycle. You'll need to use the equations that link temperature and volume that are derived using the equations for an adiabatic process.

In using the Carnot cycle as a refrigerator we cool by the removal of $Q_{L}$ from the low temperature reservoir.

The first part of this question requires us to first find an amount of heat to freeze water in to ice. To do this we need to use the specific heat of water (4186 J/kg/K), and ice (2100 J/kg/K) and also the latent heat of fusion for water (333 kJ/kg).

We then make use of the fact that

$\frac{Q_{L}}{Q_{H}}=\frac{T_{L}}{T_{H}}$ and $Q_{H}=W+Q_{L}$ to find the amount of work that must be done.

The second part of the problem requires a similar approach, but less heat is require as you only freezing the water to $0^{o}\mathrm{C}$. Once you have the work it's pretty easy to work out the time when you know the power.

In this problem the gas expands isothermally, so we can use

$\Delta S =\frac{Q}{T}$

Our main job is to find the heat in an isothermal expansion. This is the same as the work done by the gas in the expansion. You'll need to use the ideal gas law to get the volume from the given pressure.

This is a mixing problem. The first part requires you use specific heats of the two objects coming to equal temperature to find the final temperature.

The second part requires you to find the entropy change of the system.

Each object will have an entropy change:

$\Delta S_{object}=\int \frac{dQ}{T}=mc\int_{T_{initial}}^{T_{final}}\frac{dT}{T}=mc\ln\frac{T_{final}}{T_{initial}}$

find the entropy change of each object and sum them for the total entropy.