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

We are given a Carnot engine with efficiency $$\eta_1 = \frac{1}{3}$$ and a hot reservoir temperature $$T_H = 99°C = 372\text{ K}$$, and since the efficiency of a Carnot engine is given by $$\eta = 1 - \frac{T_C}{T_H}$$, substituting $$\eta = \frac{1}{3}$$ and $$T_H = 372\text{ K}$$ leads to $$\frac{1}{3} = 1 - \frac{T_C}{372}$$. Rearranging gives $$\frac{T_C}{372} = \frac{2}{3}$$, so $$T_C = \frac{2}{3} \times 372 = 248\text{ K}$$.

When the cold reservoir temperature is raised by an amount $$x$$, the new temperature becomes $$248 + x$$ and the efficiency drops to $$\frac{1}{6}$$, so substituting into the Carnot formula gives $$\frac{1}{6} = 1 - \frac{248 + x}{372}$$, which rearranges to $$\frac{248 + x}{372} = \frac{5}{6}$$, thus $$248 + x = 372 \times \frac{5}{6} = 310$$, and therefore $$x = 310 - 248 = 62\text{ K}$$.

Therefore, the required increase in the cold reservoir temperature is $$62\text{ K}$$.