Question

A Carnot heat engine operates between two reservoirs of temperatures 900 \(^{\circ}\)C (T\(_H\)) and 30 \(^{\circ}\)C (T\(_L\)). If the heat transferred during one cycle to the engine from T\(_H\) is 150 kJ, then the energy rejected to T\(_L\) is .................... kJ (round off to the nearest integer)

Hint:

Always, always convert temperatures to an absolute scale (Kelvin or Rankine) when working with thermodynamic ratios like the Carnot efficiency or the \(Q/T\) relationship. Using Celsius or Fahrenheit will give a completely wrong answer.

Answer 1

Step 1: Understanding the Concept:
The problem deals with a Carnot heat engine, which is a theoretical, reversible heat engine operating between two temperature reservoirs. The key property of a Carnot cycle is that the ratio of heat transfers is equal to the ratio of the absolute temperatures of the reservoirs.
Step 2: Key Formula or Approach:
For a Carnot (reversible) cycle, the relationship between the heat absorbed from the high-temperature reservoir (\(Q_H\)), the heat rejected to the low-temperature reservoir (\(Q_L\)), and the absolute temperatures of the reservoirs (\(T_H\) and \(T_L\)) is given by: \[ \frac{Q_L}{Q_H} = \frac{T_L}{T_H} \] It is crucial to use absolute temperatures (in Kelvin) in this formula. We can rearrange this to solve for \(Q_L\).
Step 3: Detailed Calculation:
Given values:
- High temperature, \(T_H = 900 \,^{\circ}\text{C}\)
- Low temperature, \(T_L = 30 \,^{\circ}\text{C}\)
- Heat absorbed, \(Q_H = 150\) kJ
First, convert the temperatures to Kelvin:
\[ T_H = 900 + 273.15 = 1173.15 \text{ K} \] \[ T_L = 30 + 273.15 = 303.15 \text{ K} \] Now, use the Carnot ratio to find \(Q_L\): \[ Q_L = Q_H \times \frac{T_L}{T_H} \] \[ Q_L = 150 \text{ kJ} \times \frac{303.15}{1173.15} \] \[ Q_L \approx 150 \times 0.25841 \] \[ Q_L \approx 38.76 \text{ kJ} \] Rounding off to the nearest integer: \[ Q_L = 39 \text{ kJ} \] Step 4: Final Answer:
The energy rejected to T\(_L\) is 39 kJ.