Two identical metal blocks, each of heat capacity C=1.5 kJ K^-1 (independent of temperature), are at 400 K and 100 K. A reversible heat engine…

For a reversible process the entropy of the two blocks together is unchanged: $C\ln\dfrac{T_f}{400}+C\ln\dfrac{T_f}{100}=0$, so $T_f=\sqrt{400\times100}=200\ \mathrm{K}$, not the contact-mixing value $250\ \mathrm{K}$: (A) incorrect. Heat drawn from the hot block: $Q_h=C(400-200)=300\ \mathrm{kJ}$. Heat delivered to the cold block: $Q_c=C(200-100)=150\ \mathrm{kJ}$. Work $W=Q_h-Q_c=150\ \mathrm{kJ}$ (equivalently $C(T_1+T_2-2\sqrt{T_1T_2})$): (B) correct, (C) incorrect since $300\ \mathrm{kJ}$ is $Q_h$. Overall efficiency $=150/300=50\%$; the Carnot efficiency $75\%$ holds only at the start, and it falls as the temperature difference shrinks: (D) incorrect.