Question

An engine has an efficiency of $\dfrac{1}{6}$. When the temperature of the sink is reduced by ${{62}^{0}}C$, its efficiency is doubled. The temperature of source will be –

& \text{A) 3}{{\text{7}}^{\text{0}}}\text{C} \\\ & \text{B) 6}{{\text{2}}^{\text{0}}}\text{C} \\\ & \text{C) 9}{{\text{9}}^{\text{0}}}\text{C} \\\ & \text{D) 12}{{\text{4}}^{\text{0}}}\text{C} \\\ \end{aligned}$$

Answer 1

We can easily determine the temperature of the source from the given information. We can employ our knowledge on the relation between the efficiency and temperature of the source and the sink to find the unknown temperature.

Complete step by step answer:
We know that any engine working between a source and a sink has an efficiency depending on the work done by the engine. We also know that there is no possibility of a $\text{100 }\\!\\!\%\\!\\!\text{ }$ efficient system which can be developed in nature. Every system has its own energy loss which cannot be rectified at a viable temperature difference.
We know from our past experience that the efficiency of any system is the ratio of its output to the input. In thermodynamic point of view, it is the ratio of the work done by the engine to the input energy given by the source.
i.e.,

& \text{Efficiency,}\eta =\dfrac{\text{Work done by the engine(Output),(}{{Q}_{in}}-{{Q}_{out}})}{\text{Input energy, }{{Q}_{in}}} \\\ & \Rightarrow \text{ }\eta =\dfrac{{{\text{Q}}_{in}}-{{\text{Q}}_{out}}}{{{\text{Q}}_{in}}}=1-\dfrac{{{\text{Q}}_{out}}}{{{\text{Q}}_{in}}} \\\ \end{aligned}$$ Now, we know that the temperature of the source and the sink are directly proportional to the heat at the source and sink respectively. Therefore, we can rewrite the formula for efficiency in terms of temperature as – $$\eta =\dfrac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}}=1-\dfrac{{{T}_{2}}}{{{T}_{1}}}$$ Now, let us substitute the values of temperature as per the given question. It is given that initially the efficiency was $$\dfrac{1}{6}$$. Later on, when the temperature of the sink was reduced by $${{62}^{0}}C$$, the efficiency doubles, i.e., it becomes $$\dfrac{2}{6}$$. Let us consider the case I, $$\begin{aligned} & {{\eta }_{1}}=1-\dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{1}{6} \\\ & \Rightarrow \text{ }\dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{5}{6}\text{ } \\\ & \Rightarrow \text{ }6{{T}_{2}}-5{{T}_{1}}=0\text{ ------(1)} \\\ \end{aligned}$$ Now, let us consider the case II, $$\begin{aligned} & {{\eta }_{2}}=1-\dfrac{{{T}_{2}}-62}{{{T}_{1}}}=\dfrac{2}{6}=\dfrac{1}{3} \\\ & \Rightarrow \text{ }\dfrac{{{T}_{2}}-62}{{{T}_{1}}}=\dfrac{2}{3} \\\ & \Rightarrow \text{ }3{{T}_{2}}-2{{T}_{1}}-186=0\text{ ------(2)} \\\ \end{aligned}$$ Comparing (1) and (2), we get, $$\begin{aligned} & 6{{T}_{2}}-5{{T}_{1}}=0 \\\ & 3{{T}_{2}}-2{{T}_{1}}=186 \\\ & \Rightarrow \text{ }{{T}_{1}}=372K \\\ & \text{and,} \\\ & \Rightarrow \text{ }{{T}_{2}}=310K \\\ \end{aligned}$$ We need to find the initial source temperature, which is given by $${{T}_{1}}$$. $$\Rightarrow \text{ }{{T}_{1}}=372K={{99}^{0}}C$$ **The correct answer is given by option C.** **Note:** The ratio of temperature can be taken on any scale, as the ratio is independent of the dimension of the quantity. But care should be taken that the answer which we get from the efficiency is always in the absolute scale and needs to be converted to the required scale as we did here.