A Carnot engine operating between two reservoirs has efficiency $$\frac{1}{3}$$. When the temperature of cold reservoir raised by $$x$$, its efficiency decreases to $$\frac{1}{6}$$. The value of $$x$$, if the temperature of hot reservoir is $$99°$$ C, will be

A Carnot engine operates between two reservoirs with efficiency $$\frac{1}{3}$$. When the cold reservoir temperature is raised by $$x$$, the efficiency decreases to $$\frac{1}{6}$$. The hot reservoir temperature is $$99°$$C.
In Kelvin this gives $$T_H = 99 + 273 = 372$$ K.

The initial efficiency satisfies the relation $$\eta = 1 - \frac{T_C}{T_H}$$. Substituting $$\frac{1}{3} = 1 - \frac{T_C}{372}$$ leads to $$\frac{T_C}{372} = \frac{2}{3}$$ and hence $$T_C = 248$$ K.

After increasing the cold reservoir temperature by $$x$$, the efficiency becomes $$\frac{1}{6} = 1 - \frac{T_C + x}{372}$$. Rearranging gives $$\frac{T_C + x}{372} = \frac{5}{6}$$, so $$T_C + x = 310$$.

Solving for $$x$$ yields $$x = 310 - 248 = 62$$ K, so the answer is Option D: 62 K.