Question

A thermodynamic engine operates cyclically and reversibly between two temperature reservoirs, absorbing heat from the high-temperature bath

at 450Kand discharging heat to the low-temperature bath at 300K.

(a) What is the thermodynamic efficiency of the engine?

(b) How much heat is discarded to the low-temperature bath if 1500Jof heat is absorbed from the high-temperature bath during each cycle?

(c) How much work does the engine perform in one cycle of operation?

Short answer

(a)The thermodynamic efficiency$\epsilon =33\%$

(b)The quantity of heat q_CD = -1000 J.

(c)The maximum amount of work W_net = 500 J.

Step-by-step solution

Given data

The value of higher temperature T_h = 450K.

The value of lower temperature T₁= 300K

Concept of thermal efficiency

The thermal efficiency of one heat engine may differ from another. Consider the efficiency of trustworthy heat engines to gain a better understanding of this.

Calculation of thermodynamic efficiency of engine 

The efficiency of the engine is calculated with the help of the formula:

Thermodynamic efficiency of the engine,$\mathrm{\epsilon }=\frac{{\mathrm{T}}_{\mathrm{h}}-{\mathrm{T}}_1}{{\mathrm{T}}_{\mathrm{h}}}$

Put the value of the given data in the above equation.

$\begin{array}{c}\mathrm{\epsilon }=\frac{450\mathrm{K}-300\mathrm{K}}{400\mathrm{K}}\\ =\frac{150\mathrm{K}}{400\mathrm{K}}\\ \mathrm{\epsilon }=0.333\end{array}$

$\begin{array}{c}\mathrm{Converted}\mathrm{in}\mathrm{to}\mathrm{percentage}=0.333\times 100\%\\ =33\%\end{array}$

The quantity of heat discharge

The calculation of heat is shown as:

$\mathrm{\Delta U}={\mathrm{q}}_{\mathrm{AB}}+{\mathrm{q}}_{\mathrm{CD}}+{\mathrm{w}}_{\mathrm{net}}$

Where, q_CB changes in thermal energy, for cyclic process zero

heat is discharged at the lower temperature.

Now, put the value of the given data in the above equation.

$\begin{array}{c}\mathrm{\Delta U}={\mathrm{q}}_{\mathrm{AB}}+{\mathrm{q}}_{\mathrm{CD}}+{\mathrm{w}}_{\mathrm{net}}\\ 0={\mathrm{q}}_{\mathrm{AB}}+{\mathrm{q}}_{\mathrm{CD}}+{\mathrm{w}}_{\mathrm{net}}\\ {\mathrm{q}}_{\mathrm{CD}}=-{\mathrm{q}}_{\mathrm{AB}}-{\mathrm{w}}_{\mathrm{net}}\end{array}$

Put the value in the above expression.

$\begin{array}{c}{\mathrm{q}}_{\mathrm{CD}}=-1500\mathrm{J}-(-500\mathrm{J})\\ =-1000\mathrm{J}\end{array}$

 Step 4: The calculation of maximum work

The calculation of the maximum amount of work is shown with help the formula:

$\mathrm{\epsilon }=\frac{-{\mathrm{w}}_{\mathrm{net}}}{{\mathrm{q}}_{\mathrm{AB}}}$

where -W_net is work done, q_ABis heat.

Rearrange the above equation.

${\mathrm{w}}_{\mathrm{net}}=-\mathrm{\epsilon }\times {\mathrm{q}}_{\mathrm{AB}}$ …. (i)

Now, put the value of the given data in the above equation.

$\begin{array}{c}{\mathrm{w}}_{\mathrm{net}}=-0.333\times 1000\mathrm{J}\\ =-500\mathrm{J}\end{array}$