Question

An engine has a hot-reservoir temperature of 970 K and a cold-reservoir temperature of 480 K. The engine operates at three-fifths maximum efficiency. What is the efficiency of the engine?

Answer 1

Answer:

$\eta=0.303$

Explanation:

Given that

Temperature of the hot reservoir ,T₁ = 970 K

Temperature of the cold reservoir ,T₂ = 480 K

We know that only Carnot engine is the ideal engine which gives us the maximum power.The efficiency of Carnot engine is given as

$\eta_{max}=1-\dfrac{T_2}{T_1}$

$\eta_{max}=1-\dfrac{480}{970}$

$\eta_{max}=0.505$

Therefore the efficiency of the given engine will be

$\eta=\dfrac{3}{5}\eta_{max}$

$\eta=\dfrac{3}{5}\times 0.505$

$\eta=0.303$