Let _1 be the efficiency of a Carnot engine at T_H = 547^ ^ C and...

The temperatures for the first Carnot engine in Kelvin are:

$T_{H₁} = 547^{{}^{\circ}}\text{C} + 273 = 820\text{ K}$

$T_{C1} = 247^{{}^{\circ}}\text{C} + 273 = 520\text{ K}$

The efficiency $\eta_1$ is given by:

$\eta_1 = 1 - \dfrac{T_{C1}}{T_{H₁}} = 1 - \dfrac{520}{820} = 1 - \dfrac{26}{41} = \dfrac{15}{41}$

The temperatures for the second Carnot engine in Kelvin are:

$T_{H₂} = 847^{{}^{\circ}}\text{C} + 273 = 1120\text{ K}$

$T_{C2} = 147^{{}^{\circ}}\text{C} + 273 = 420\text{ K}$

The efficiency $\eta_2$ is given by:

$\eta_2 = 1 - \dfrac{T_{C2}}{T_{H₂}} = 1 - \dfrac{420}{1120} = 1 - \dfrac{3}{8} = \dfrac{5}{8}$

Given the relation $\eta_1 = \dfrac{x}{41}\eta_2$, substituting the values of $\eta_1$ and $\eta_2$:

$\dfrac{15}{41} = \dfrac{x}{41} \times \dfrac{5}{8}$

$15 = \dfrac{5x}{8}$

$x = \dfrac{15 \times 8}{5} = 24$

Answer: $24$