Question

A Carnot engine operating between temperature \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\) has efficiency \(0.4\), when \(\mathrm{T}_{2}\) lowered by $50 \mathrm{~K}\(, its efficiency increases to \)0.5\(. Then \)\mathrm{T}_{1}$ and \(\mathrm{T}_{2}\) are respectively. (A) \(300 \mathrm{~K}\) and \(100 \mathrm{~K}\) (B) \(400 \mathrm{~K}\) and \(200 \mathrm{~K}\) (C) \(600 \mathrm{~K}\) and \(400 \mathrm{~K}\) (D) \(500 \mathrm{~K}\) and \(300 \mathrm{~K}\)

Short answer

The temperatures T1 and T2 are 500 K and 300 K, respectively (Option D).

Step-by-step solution

Write down the formula for Carnot engine efficiency

For a Carnot engine operating between two temperatures T1 (higher) and T2 (lower), the efficiency is given by: Efficiency = \(1 - \frac{T_2}{T_1}\)

Equate efficiencies

Given efficiency 1 = 0.4, and efficiency 2 = 0.5. We also know that in the second case, T2 is lowered by 50 K. Efficiency 1 = \(1 - \frac{T_2}{T_1}\) = 0.4 And Efficiency 2 = \(1 - \frac{T_2 - 50}{T_1}\) = 0.5

Solve for T1 and T2

Now, we need to solve the two equations to find T1 and T2. From Efficiency 1 equation, we get: T2 = T1 (1 - 0.4) = 0.6 * T1 From Efficiency 2 equation, we get: T2 - 50 = T1 (1 - 0.5) = 0.5 * T1 Substituting T2 from the first equation into the second equation: 0.6 * T1 - 50 = 0.5 * T1 Solving for T1: 0.1 * T1 = 50 T1 = 500 K Now, we can find T2 using T1: T2 = 0.6 * T1 = 0.6 * 500 = 300 K

Compare with given options

Comparing our found values of T1 and T2 with the given options, we can conclude the correct answer is: (D) T1 = \(500 \mathrm{~K}\) and T2 = \(300 \mathrm{~K}\)