Series Carnot Engines — Finding the Intermediate Temperature Two reversible (Carnot) heat engines are arranged in series. Engine 1 operates between 1600 K (hot reservoir) and T2 K; Engine 2 operates between T2 K and 400 K. If the two engines have equal efficiencies (same heat input and output for each), determine the intermediate temperature T2.

Introduction / Context:
Carnot engines provide an ideal benchmark. When two reversible engines are placed in series with equal performance metrics (equal efficiency or identical heat input/output ratios), we can relate their temperature ratios to find the unknown intermediate temperature.

Given Data / Assumptions:

• Engine 1: T_hot1 = 1600 K, T_cold1 = T2.
• Engine 2: T_hot2 = T2, T_cold2 = 400 K.
• Both engines are reversible (Carnot) and have equal efficiencies.

Concept / Approach:
For a Carnot engine, efficiency eta = 1 - T_cold/T_hot. Equal efficiencies imply 1 - T2/1600 = 1 - 400/T2, which directly yields T2.

Step-by-Step Solution:
Set eta1 = eta2 ⇒ 1 - T2/1600 = 1 - 400/T2.Rearrange ⇒ T2/1600 = 400/T2.Multiply both sides by 1600*T2 ⇒ T2^2 = 1600 * 400 = 640000.Take positive root: T2 = sqrt(640000) = 800 K.

Verification / Alternative check:
Check efficiencies: eta1 = 1 - 800/1600 = 0.5; eta2 = 1 - 400/800 = 0.5. Both equal 50%, confirming the result.

Why Other Options Are Wrong:
1000 K, 1200 K, or 1400 K would lead to unequal efficiencies (not both 50%).

Common Pitfalls:
Equating heat transfers instead of efficiencies without a consistent basis; forgetting to use absolute temperatures.

Final Answer:
800 K