Question

A heat engine operating between energy reservoirs at 20 deg C and 620 deg C has 26 % of the maximum possible efficiency. How much energy must this engine extract from the hot reservoir to do 1200 J of work?

Answer 1

Concepts and reason

The concept required to solve the given problem is efficiency of thermal engine.

Calculate the efficiency of thermal engine and then calculate the heat extracted from the hot reservoir by taking a ratio of the work done and efficiency thus calculated.

Fundamentals

The maximum possible efficiency of a heat engine is given by,

$\eta = 1 - \frac{{{T_2}}}{{{T_1}}}$

Here, ${T_1}{\rm{ and }}{T_2}$ are the temperatures of the hot and cold reservoir respectively.

The efficiency of thermal engine can also be calculated as,

$\eta = \frac{W}{Q}$

Here, $W$ is the work done and $Q$ is the heat extracted from the hot reservoir.

The maximum possible efficiency of a heat engine is given by,

$\eta = 1 - \frac{{{T_2}}}{{{T_1}}}$ …… (1)

Here, ${T_1}{\rm{ and }}{T_2}$ are the temperatures of the hot and cold reservoir respectively.

The temperature of the hot reservoir in kelvin is given as follows:

$\begin{array}{c}\\{T_1} = \left( {620^\circ {\rm{ C}}} \right) + \left( {273.15{\rm{ K}}} \right)\\\\ = 893.15{\rm{ K}}\\\end{array}$

The temperature of the cold reservoir in kelvin is given as follows:

$\begin{array}{c}\\{T_2} = \left( {20^\circ {\rm{ C}}} \right) + \left( {273.15{\rm{ K}}} \right)\\\\ = 293.15{\rm{ K}}\\\end{array}$

Substitute $893.15{\rm{ K}}$ for ${T_1}$ and $293.15{\rm{ K}}$ for ${T_2}$ in the equation (1).

$\begin{array}{c}\\\eta = 1 - \frac{{293.15{\rm{ K}}}}{{893.15{\rm{ K}}}}\\\\ = 0.672\\\end{array}$

The efficiency of Carnot engine is,

$\begin{array}{c}\\\eta ' = \frac{{26}}{{100}} \times 0.672\\\\ = 0.175\\\end{array}$

The efficiency of thermal engine in terms of work done and heat energy extracted from hot reservoir is given as follows:

$\eta ' = \frac{W}{Q}$

Here, $W$ is the work done and $Q$ is the heat extracted from the hot reservoir.

Rearrange the above equation for Q.

$Q = \frac{W}{{\eta '}}$

Substitute $0.175$ for $\eta '$ and $1200{\rm{ J}}$ for $W$ in the above equation.

$\begin{array}{c}\\Q = \frac{{1200{\rm{ J}}}}{{0.175}}\\\\ = 6.86 \times {10^3}{\rm{ J}}\\\end{array}$

Ans:

The heat energy that must be extracted from the hot reservoir is $6.86 \times {10^3}{\rm{ J}}$ .