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

Option (1) \(2.33 \, \text{kJ/sec}\).

Step-by-step solution

- Identify Parameters

Given:- Temperature at which heat is absorbed, \(T_1 = 270 \, \text{K}\)- Temperature at which heat is rejected, \(T_2 = 300 \, \text{K}\)- Heat absorbed per minute, \(Q_1 = 1260 \, \text{kJ/min}\)

- Convert Heat Absorbed to Per Second

Since the heat absorbed is given per minute, convert it to per second:\[Q_1 \text{ (per second)} = \frac{1260 \, \text{kJ/min}}{60 \, \text{sec/min}} = 21 \, \text{kJ/sec}\]

- Calculate Coefficient of Performance (COP) for Refrigeration

The Coefficient of Performance (COP) for a Carnot refrigeration cycle is given by:\[\text{COP} = \frac{T_1}{T_2 - T_1} = \frac{270}{300 - 270} = \frac{270}{30} = 9\]

- Calculate Work Required

The work required for the refrigeration cycle is given by:\[W = \frac{Q_1}{\text{COP}} = \frac{21 \, \text{kJ/sec}}{9} = 2.33 \, \text{kJ/sec}\]

- Select the Correct Option

The correct option based on the calculated work required is:\(2.33 \, \text{kJ/sec}\). Therefore, option (1) is correct.

Key concepts

Coefficient of Performance (COP)

The Coefficient of Performance (COP) is a crucial concept in refrigeration cycles. It measures the efficiency of a refrigeration system. The higher the COP, the more efficient the system is. For a Carnot refrigeration cycle, the COP is calculated by the formula:
\(\text{COP} = \frac{T_1}{T_2 - T_1}\)
where \( T_1 \) is the temperature at which heat is absorbed, and \( T_2 \) is the temperature at which heat is rejected. In this problem:
- \( T_1 = 270 \) Kelvin
- \( T_2 = 300 \) Kelvin
Using the values, we find:
\( \text{COP} = \frac{270}{300 - 270} = \frac{270}{30} = 9 \).
This tells us that the refrigeration cycle can transfer 9 units of heat for each unit of work done. The COP is a dimensionless number, and a higher COP indicates a more efficient refrigeration cycle.

Thermodynamics

Thermodynamics is the branch of physics that deals with heat and temperature, and their relation to energy and work. One of the key applications of thermodynamics is in understanding refrigeration cycles like the Carnot cycle.
A Carnot refrigeration cycle operates between two temperatures: a cold temperature \( T_1 \) at which it absorbs heat and a higher temperature \( T_2 \) where it rejects heat. The difference in these temperatures drives the cycle and determines its efficiency.
The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted from one form to another. For refrigeration cycles, this principle is used to calculate the work required and heat transferred during the process. The second law of thermodynamics tells us that heat cannot spontaneously flow from a colder area to a hotter area; work must be done to achieve this, which is the primary function of refrigeration systems.

Work calculation in refrigeration cycles

In a refrigeration cycle, work is required to transfer heat from a lower temperature region to a higher temperature one. Calculating this work involves understanding the relationship between heat absorbed, the COP, and the temperatures involved.
Using the given problem, heat absorbed per second by the system was calculated as follows:
- Given: 1260 kJ/min
- Conversion: \( \frac{1260 \text{kJ/min}}{60 \text{sec/min}} = 21 \text{kJ/sec} \)
With the COP known to be 9, the work required (W) is calculated by the formula:
\( W = \frac{Q_1}{\text{COP}} \)
Substituting the values:
\( W = \frac{21 \text{kJ/sec}}{9} = 2.33 \text{kJ/sec} \)
This shows that the refrigeration cycle requires 2.33 kJ of work per second to move 21 kJ of heat per second from the lower temperature reservoir to the higher one. This calculation is essential in designing and analyzing refrigeration and air conditioning systems for efficiency and performance.