Thermodynamics Question - Entropy etc

[doubleB]

Homework Statement

I've attached a J-peg with the question in, laid out much more clearly than I could expect to type it here. I hope the attachment is ok.

Homework Equations

The Attempt at a Solution

So far I have:

(1) dS = dQ / T
Entropy is a measure of a system's disorder.

(2) I've drawn a flow diagram but can't get a picture of that on here. Suffice to say I have tried to do this part quite hard but can't get it to work.

(3)

(a) dS = dQ /T ... dQ = 3/2 N K dT ... so dS = 3/2 N K dT/T. Then I can work out the integral and do this first part for temperature going from 293 to 273? Does mass not matter here?

(b) dS = dQ/T ... We know Q from the question as 333kJ/kg so S = Q/T goes to S = 333 * 1 kilogram of water / 273 Kelvin. Is this right?

(c) I have no idea how to do this.

The entropy change of the universe will increase during this process, as it will do for all processes above 0 Kelvin.

(4) Coefficient of Performance = Qh / Work = Qh / Qh - Qc ... this can then be written as Th / Th - Tc = 323K / 323K - 273K = 6.46.

To improve the coefficient of performance of the ice-maker you could decrease the temperature of the hot reservoir? This doesn't seem to make sense...

If anyone has the time to look at this and help me I would very much appreciate it as I have lost all my second year notes on thermodynamics and can't see how to do this question properly.

DoubleB

 

[doubleB]

Can anyone help me?

 

[m2003]

Dear DoubleB,

Thank you for reaching out for help with this thermodynamics question. I will do my best to guide you through the problem and provide some insights along the way.

Firstly, you have correctly identified that entropy is a measure of a system's disorder. It is a fundamental concept in thermodynamics and plays a crucial role in understanding the behavior of physical systems.

Now let's move on to your attempt at solving the problem. For part (a), you are on the right track by using the equation dS = dQ/T. However, you also need to consider the fact that entropy is an extensive property, meaning it depends on the amount of material you have. In this case, mass does matter and you will need to include it in your calculation. The correct expression for the change in entropy for an ideal gas is given by dS = N C_v ln(T2/T1), where N is the number of particles and C_v is the heat capacity at constant volume. You can use this equation to calculate the change in entropy for the temperature range given in the problem.

For part (b), you are correct in using the equation dS = dQ/T. However, you need to be careful with your units. The heat capacity given in the problem is in kJ/kg, so you will need to convert it to kJ/kg-K before using it in the equation. Also, keep in mind that the change in entropy for this part will be for the water, not the ice.

For part (c), you will need to use the second law of thermodynamics, which states that the total entropy of a closed system can never decrease. This means that the entropy change of the universe must be greater than or equal to zero. For this part, you will need to calculate the entropy change of the universe by considering the entropy changes of the hot and cold reservoirs.

Finally, for part (4), you are correct in using the equation for the coefficient of performance, but you should also consider the Carnot efficiency, which is given by Th/(Th-Tc). This represents the maximum theoretical efficiency for a heat engine operating between two temperatures. By decreasing the temperature of the hot reservoir, you can increase the Carnot efficiency and therefore improve the coefficient of performance.

I hope this helps you in solving the problem. If you have any further questions, please don't hesitate to ask. Keep up the good work and don't get discouraged by the complexity of