Question

A Carnot refrigerator, working between (0°C) and (30°C), is used to cool a bucket of water containing (10⁻² m³) of water at (30°C) to (5°C) in (2) hours. Find the total amount of work needed.

a) Calculate the total work needed.
b) Determine the efficiency of the Carnot refrigerator.

Answer 1

The total work required by a Carnot refrigerator to cool 10-2 m^3 of water from 30°C to 5°C is approximately 11628 J, and the efficiency is about 9.9%.

Step-by-step explanation:

To find the total amount of work needed by a Carnot refrigerator to cool a bucket of water from 30°C to 5°C, we can use the concept of the coefficient of performance (COP) of a Carnot refrigerator, which is defined as:

COP = Qc / W = Tc / (Th - Tc)

First, we need to calculate the heat removed from the water:

Qc = mass × specific heat capacity of water × change in temperature

Qc = (10-2 m3 × 1000 kg/m3) × 4186 J/kg·K × (30°C - 5°C)

Qc = 104650 J

Next, we calculate the COP for the temperatures given in Kelvin (273 + given °C):

COP = 273 K / (303 K - 273 K) = 9

Now, we can find the total work (W) needed:

W = Qc / COP = 104650 J / 9 ≈ 11628 J

To determine the efficiency of the Carnot refrigerator, we use the formula:

e = 1 - (Tc / Th)

e = 1 - (273 K / 303 K) ≈ 0.099

The efficiency of the Carnot refrigerator is approximately 9.9%.