Question

An ideal gas heat engine operates in the Carnot cycle between $227^o C$ and $127 ^o C$. It absorbs 6 $\times 10 ^4$ cal of heat. The amount of heat converted into work is equal to:
$\begin{aligned} & A.2.4\times {{10}^{4}}cal \\\ & B.6\times {{10}^{4}}cal \\\ & C.1.2\times {{10}^{4}}cal \\\ & D.4.8\times {{10}^{4}}cal \\\ \end{aligned}$

Answer 1

Hint: Use the formula of efficiency of a heat engine in terms of temperatures. Using the formula of efficiency of a heat engine in terms of work done and heat absorbed, calculate the value of the work done.

Step by step solution:
The given temperatures of operation of the Carnot cycle are $227^o C$ and $127^o C$.
Here we need to convert the given temperatures into Absolute Temperature scale which is the Kelvin scale.
We know conversion between the Kelvin scale and centigrade scale can be done as
Kelvin scale temperature = centigrade scale + 273
So we get the value of temperature in terms of the Kelvin scale is:
$T_1 = 273 + 227 = 500 K$
$T_2 = 273 + 127 = 400 K$
Having got the temperatures of the Carnot cycle in terms of the Kelvin scale, we can now calculate the efficiency:
We know the efficiency of the heat engine is given by
$\eta = 1 -\dfrac{T_2}{T_1}$
Also, we can write the efficiency in terms of work done and heat absorbed as
$\eta = \dfrac{W}{Q}$
Here Q is the amount of heat observed from the source of heat and $T_1$ is the temperature of the source and $T_2$ is that temperature of the sink and W is the amount of work
So we get using the above two expressions as :
$\eta = 1 - \dfrac{T_2}{T_1} = 1 -\dfrac{400}{500} = \dfrac{1}{5}$
Also we get: $\eta = \dfrac{W}{Q} = \dfrac{W}{6 \times 10^4}$
$\dfrac{W}{6 \times 10^4} = \dfrac{1}{5}$
We get $W = \dfrac{6 \times 10^4}{5} = 1.2 \times 10^4$
Thus using the efficiency of heat engine formula in terms of both temperature and formula in terms of work and heat energy to calculate work done as $1.2 \times 10^4$ calories.

Note: The possible mistakes that one can make in this kind of problem is the confusion between the temperatures $T_1$ and $T_2$. We need to take care of the definitions of temperatures $T_1$ and $T_2$. Note that $T_2$ refers to the temperature of the sink and $T_1$ refers to the temperature of the source.