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(B) The efficiency of a Carnot heat engine is given by $\eta = 1 - \frac{T_{sink}}{T_{source}}$.Given that the two heat engines $X$ and $Y$ have the s

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$600$

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(B) The efficiency of a Carnot heat engine is given by $\eta = 1 - \frac{T_{sink}}{T_{source}}$.Given that the two heat engines $X$ and $Y$ have the same efficiency, $\eta_X = \eta_Y$.For engine $X$, the source temperature is $900 \,K$ and the sink temperature is $T \,K$. Thus, $\eta_X = 1 - \frac{T}{900}$.For engine $Y$, the source temperature is $T \,K$ and the sink temperature is $400 \,K$. Thus, $\eta_Y = 1 - \frac{400}{T}$.Equating the efficiencies:$1 - \frac{T}{900} = 1 - \frac{400}{T}$$\frac{T}{900} = \frac{400}{T}$$T^2 = 900 	imes 400$$T^2 = 360000$$T = \sqrt{360000} = 600 \,K$.Therefore, the temperature $T$ is $600 \,K$.

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