Thermodynamics/Statisical Mechanics: Carnot Cycle

[Seda]

Homework Statement

Consider a Carnot Cycle that absorbs heat from the hot reservoir at T_hw and expels to the cold reservoir at T_cw just below T_h and above T_c respectively.

a.) Derive an equation that relates the four temperatures.

b.) Using the first and second law, write down an expression for power entirely in terms of the four temperatures (and the constant K) and use the results of (a) to eliminate T_cw.

c.) Show that the expression found in (b) has a maximum value at T_hw=.5(T_h + sqrt[T_hT_c]). Find a corresponding expression for T_cw.

d.) Show that the efficiency of the engine is 1-sqrt(T_c/T_h)

Homework Equations

Q_h/t = K(T_h-T_hw)
Q_c/t = K(T_cw-T_c)

K is the proportionality constant are the same for both processes.

The Attempt at a Solution

a.) I solved this knowing that the cycle's change in entropy is zero using S=Q/T.

My answer for a is: Q_h[(1/T_hw)-(1/T_h)]+Q_c[(1/T_c)-(1/T_cw)]=0

b.) This is where I'm stuck. I can solve the equation above for T_cw=1/[(Q_h/Q_c)[(1/T_hw)-(1/T_h)]+(1/T_c)]

I think I can substitute this monster above into this equation for power for T_cw

P=K(T_h-T_hw) - K(T_cw-T_c)

to solve part (b).

However, once i start taking the derivative of this to find a max for part c, the math gets ridiculous, I believe I have made an error or can somehow simplify this before attempting part c. Any tips?

 

[mark726]

Dear student,

Thank you for your post. Let's work through this problem together.

a.) To derive the equation relating the four temperatures, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In a Carnot cycle, the work done is equal to the difference between the heat absorbed from the hot reservoir and the heat expelled to the cold reservoir, so we have:

ΔU = Qh - Qc

We also know from the second law of thermodynamics that the efficiency of a Carnot cycle is given by:

η = 1 - Qc/Qh

We can rearrange this equation to solve for Qc:

Qc = Qh - ηQh = Qh(1 - η)

Substituting this into our first equation, we get:

ΔU = Qh - Qh(1 - η) = ηQh

We also know that the change in internal energy is equal to the heat added to the system (Qh) minus the work done by the system (W), so we can write:

ΔU = Qh - W

Substituting this into our equation for ΔU, we get:

ηQh = Qh - W

Solving for W, we get:

W = (1 - η)Qh

Now, we can use the definition of work (W = PΔV) and the ideal gas law (PV = nRT) to write:

W = (1 - η)Qh = (1 - η)nRT

We can also use the definition of efficiency (η = Qh/Qc) to write:

Qc = Qh - ηQh = Qh(1 - η)

Substituting this into our equation for work, we get:

W = (1 - η)Qh = (1 - η)Qc/(1 - η) = Qc

Now, we can use the definition of heat (Q = mcΔT) to write:

Qh = mcΔTh

Qc = mcΔTc

Substituting these into our equation for work, we get:

W = Qc = (1 - η)mc(ΔTc - ΔTh)

Now,