Question

A carnot refrigeration cycle absorbs heat at \(270 \mathrm{~K}\) and rejects heat at \(300 \mathrm{~K}\). If the cycle is absorbing \(1260 \mathrm{KJ} / \mathrm{min}\) at \(270 \mathrm{~K}\), then work required per second is: (1) \(2.33 \mathrm{KJ} / \mathrm{sec}\). (2) \(4.66 \mathrm{KJ} / \mathrm{sec}\). (3) \(1 \mathrm{KJ} / \mathrm{sec}\). (4) \(4 \mathrm{KJ} / \mathrm{sec} .\)

Short answer

The work required per second is \ 2.33 \mathrm{kJ/s} \. Option (1) is correct.

Step-by-step solution

Understand the given data

Given: - Temperature of the cold reservoir, \(T_{C} = 270 \mathrm{K}\). - Temperature of the hot reservoir, \(T_{H} = 300 \mathrm{K}\). - Heat absorbed per minute, \(Q_{C} = 1260 \mathrm{kJ/min}\).

Convert heat absorbed to per second

Convert heat absorbed from per minute to per second: \[Q_{C} \text{(per second)} = \frac{1260 \mathrm{kJ}}{60 \mathrm{s}} = 21 \mathrm{kJ/s}.\]

Write the Carnot refrigeration efficiency formula

The coefficient of performance (COP) for a Carnot refrigeration cycle is given by: \[COP = \frac{T_C}{T_H - T_C} = \frac{270}{300 - 270} = \frac{270}{30} = 9.\]

Find the work input requirement

The formula relating the heat absorbed in the cold reservoir, the work done, and the COP is: \[COP = \frac{Q_{C}}{W}.\] Rearranging for work: \[W = \frac{Q_{C}}{COP} = \frac{21 \mathrm{kJ/s}}{9} = 2.33 \mathrm{kJ/s}.\]

Verify the answer

By comparing to the provided options, the correct answer matches with option (1) \(2.33 \mathrm{kJ/s}\).

Key concepts

Thermodynamics

Thermodynamics is the study of energy transfer, particularly heat and work, and how it affects matter. In the context of a Carnot refrigeration cycle, the thermodynamic principles help us determine the work input required for the cycle. The first law of thermodynamics, also known as the conservation of energy, states that energy cannot be created or destroyed, only transformed. In a refrigeration cycle, energy is transferred from a cooler region (cold reservoir) to a warmer region (hot reservoir), requiring work input to achieve this flow against the natural heat gradient.

Heat Transfer

Heat transfer is the process of energy movement in the form of heat from one place to another because of temperature differences. In the Carnot refrigeration cycle, heat transfer takes place between two reservoirs at different temperatures. The cold reservoir, where the heat is absorbed, is at 270 K, and the hot reservoir, where the heat is rejected, is at 300 K. The efficiency of this transfer is crucial for determining the work input needed. For the Carnot cycle, the heat absorbed (Q_C) per second is calculated by converting the given value from minutes to seconds.

• Heat absorbed per minute: 1260 kJ/min.
• Convert to per second: Q_C (per second) = 1260 kJ / 60 s = 21 kJ/s.

Coefficient of Performance (COP)

The coefficient of performance (COP) measures the efficiency of a refrigeration cycle. It is the ratio of heat absorbed from the cold reservoir to the work input required. In mathematical terms, for a Carnot refrigeration cycle, COP is given by:
\[ COP = \frac{T_C}{T_H - T_C} \]

• Where T_C is the temperature of the cold reservoir (270 K).
• T_H is the temperature of the hot reservoir (300 K).

In our example, the COP is calculated as:
\[ COP = \frac{270}{300 - 270} = \frac{270}{30} = 9 \]
The work input (W) is then derived from the COP using the formula:
\[ W = \frac{Q_C}{COP} \]
Calculating this for our given values:
\[ W = \frac{21 \text{kJ/s}}{9} = 2.33 \text{kJ/s} \]
This means option (1) 2.33 kJ/s is the correct amount of work required per second.