Question

A carnot engine works between the temperatures $327^\circ\text{C}$ and $227^\circ\text{C}$. If the work output of the engine is $1\text{ kJ}$, then the amount of heat absorbed by the engine is

• A. 2 kJ
• B. 6 kJ
• C. 30 kJ
• D. 3 kJ
• E. 40 kJ

Hint:

Always convert temperatures to Kelvin in thermodynamics problems. Using Celsius will lead to incorrect efficiency values.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The efficiency ($\eta$) of a Carnot engine depends only on the temperatures of the hot reservoir ($T_1$) and cold reservoir ($T_2$). It is also defined as the ratio of work output ($W$) to heat absorbed ($Q_1$).
Key Formula or Approach:
1. Convert temperatures to Kelvin: $T(K) = T(^\circ\text{C}) + 273$.
2. Efficiency $\eta = 1 - \frac{T_2}{T_1}$.
3. Heat absorbed $Q_1 = \frac{W}{\eta}$.

Step 2: Detailed Explanation:
1. Convert temperatures:
$T_1 = 327 + 273 = 600\text{ K}$ (Source temperature)
$T_2 = 227 + 273 = 500\text{ K}$ (Sink temperature)
2. Calculate efficiency:
\[ \eta = 1 - \frac{500}{600} = 1 - \frac{5}{6} = \frac{1}{6} \]
3. Find heat absorbed $Q_1$:
Given work output $W = 1\text{ kJ}$.
\[ \eta = \frac{W}{Q_1} \implies \frac{1}{6} = \frac{1\text{ kJ}}{Q_1} \]
\[ Q_1 = 6\text{ kJ} \]

Step 3: Final Answer:
The amount of heat absorbed by the engine is 6 kJ.