91Ó°ÊÓ

A Carnot cycle on the T-S diagram with the scale set from the graph,

${\mathrm{S}}_{\mathrm{a}}\left({\mathrm{S}}_{\mathrm{c}}\right)=0.10\mathrm{J}/\mathrm{K},{\mathrm{T}}_{\mathrm{H}}=400\mathrm{K}\mathrm{and}{\mathrm{S}}_{\mathrm{b}}\left({\mathrm{S}}_{\mathrm{d}}\right)=600\mathrm{J}/\mathrm{K}$

Equation 20-1 tells us that any energy transfer as the heat must involve a change in entropy. From the figure, at the vertical lines, entropy is constant. From the second law of thermodynamics,$\mathbf{∆}\mathbf{S}\mathbf{=}\mathbf{∫}\mathbf{dQ}\mathbf{/}\mathbf{T}$ .So, the heat is also constant at vertical lines. That means the process involved at this point is adiabatic. Similarly, the two horizontal lines correspond to two isothermal processes of the Carnot cycle i.e. temperature is constant. For simplification, we can redraw the figure and find the total heat transformed and net work done by the system.

Formulae:

The work done per cycle due to the first law of thermodynamics,

$\mathrm{W}=\left|{\mathrm{Q}}_{\mathrm{H}}\right|-\left|{\mathrm{Q}}_{\mathrm{L}}\right|$ (1)

The entropy change of the cycle using the second law of thermodynamics,

$∆\mathrm{S}=∫\frac{\mathrm{dQ}}{\mathrm{T}}$ (2)

$$\begin{gathered} {\mathrm{dQ}}_{\mathrm{L}}=∆\mathrm{S}×{\mathrm{T}}_{\mathrm{L}} \\ =0.50\mathrm{J}/\mathrm{K}×250\mathrm{K} \\ =75\mathrm{J} \end{gathered}$$For simplification, we redraw the figure.

For paths ab and cd , the process is isothermal, so the temperature is constant. Let ${\mathrm{Q}}_{\mathrm{H}}$be the heat transformed at the path ab and ${\mathrm{Q}}_{\mathrm{L}}$be the heat transformed at the path cd . For paths ac and bc, the process is adiabatic, so the total heat is constant and entropy is also constant.

As entropy is set by the scale${\mathrm{S}}_{\mathrm{s}}=0.60\mathrm{J}/\mathrm{K}$

The change in entropy for the pathis given as:

$$\begin{gathered} ∆\mathrm{S}={\mathrm{S}}_{\mathrm{b}}-{\mathrm{S}}_{\mathrm{a}} \\ =0.60\mathrm{J}/\mathrm{k}-0.10\mathrm{J}/\mathrm{K} \\ =0.50\mathrm{J}/\mathrm{k} \end{gathered}$$

The net heat transformed at high temperature,i.e., ${\mathrm{T}}_{\mathrm{H}}=400\mathrm{K}$ is calculated by using equation (2) as:

role="math" localid="1661337709961" $$\begin{gathered} {\mathrm{dQ}}_{\mathrm{H}}=∆\mathrm{S}×{\mathrm{T}}_{\mathrm{H}} \\ =0.50\mathrm{J}/\mathrm{K}×400\mathrm{K} \\ =200\mathrm{J} \end{gathered}$$

The cycle is complete, so the total internal energy is constant.

Similarly, for path cd, the change in entropy is given as:

$$\begin{gathered} ∆\mathrm{S}={\mathrm{S}}_{\mathrm{c}}-{\mathrm{S}}_{\mathrm{d}} \\ =0.10\mathrm{J}/\mathrm{K}-0.60\mathrm{J}/\mathrm{k} \\ =-0.50\mathrm{J}/\mathrm{K} \end{gathered}$$

The net heat transformed at low temperature i.e.${\mathrm{T}}_{\mathrm{L}}=250\mathrm{K}$is calculated by using equation (2) as:

$$\begin{gathered} {\mathrm{dQ}}_L=∆\mathrm{S}×{\mathrm{T}}_L \\ =0.50\mathrm{J}/\mathrm{K}×250\mathrm{K} \\ =125\mathrm{J} \end{gathered}$$

Now, the net heat transferred in the cycle is given as:

$$\begin{gathered} {\mathrm{dQ}}_{\mathrm{H}}={\mathrm{dQ}}_{\mathrm{H}}+{\mathrm{dQ}}_{\mathrm{L}} \\ =200\mathrm{J}-125\mathrm{J} \\ =75\mathrm{J} \end{gathered}$$

Hence, the value of the net energy transfer is 75 J

According to the first law of thermodynamics,

$\mathrm{dQ}=∆{\mathrm{E}}_{\mathrm{int}}+\mathrm{dW}$

For the complete cycle, change in internal energy is zero i.e.$∆{\mathrm{E}}_{\mathrm{int}}=0\mathrm{J}$

So, the total heat change is the work done by the system, dQ = dW, i.e.,

dW = 75 J

Hence, the value of the net work done is 75 J