Question

Efficiency of a Carnot engine is \(50 \%\), when temperature of outlet is $500 \mathrm{~K}\(. in order to increase efficiency up to \)60 \%$ keeping temperature of intake the same what is temperature of out let. (A) \(200 \mathrm{~K}\) (B) \(400 \mathrm{~K}\) (C) \(600 \mathrm{~K}\) (D) \(800 \mathrm{~K}\)

Short answer

The new temperature of the outlet to have an efficiency of 60% is 400 K, so the answer is (B) \(400\,\mathrm{K}\).

Step-by-step solution

Write down the Carnot efficiency formula

The efficiency of a Carnot engine is given by: \( \eta = 1 - \frac{T_{outlet}}{T_{intake}} \) Where: - \( \eta \) is the efficiency of the engine - \( T_{outlet} \) is the temperature of the outlet - \( T_{intake} \) is the temperature of the intake

Find the temperature of the intake using the given efficiency and outlet temperature

We know the initial efficiency is 50% and the outlet temperature is 500 K. We must find the temperature of the intake using the formula above. \( 0.50 = 1 - \frac{500}{T_{intake}} \) First, we will solve for \( T_{intake} \): \( 0.50 = 1 - \frac{500}{T_{intake}} \Rightarrow T_{intake} = \frac{500}{0.50}\) \( T_{intake} = 1000\,K \)

Find the new outlet temperature for the desired efficiency

Now that we have the intake temperature, we want to find the new outlet temperature for an efficiency of 60%. We will use the efficiency formula again with the new efficiency value: \( 0.60 = 1 - \frac{T_{outlet_{new}}}{1000} \) Now solve for \( T_{outlet_{new}} \): \( 0.60 = 1 - \frac{T_{outlet_{new}}}{1000} \Rightarrow T_{outlet_{new}} = 1000 - (1000\times0.60) \) \( T_{outlet_{new}} = 1000 - 600 \) \( T_{outlet_{new}} = 400\,K \)

Choose the correct answer from the given options

The new temperature of the outlet to have an efficiency of 60% is 400 K, so the answer is (B) \(400\,\mathrm{K}\).