Carnot Refrigerator coefficient of performance

[Kalie]

A Carnot refrigerator operating between -10.0 C and 40.0 C extracts heat from the cold reservoir at the rate 300 J/s. What are (a) the coefficient of performance of this refrigerator, (b) the rate at which work is done on the refrigerator and (c) the rate at which heat is exhausted to the hot side?

A. All right for i got the answer to be 5.26 because
T_c/(T_h-T_c)=coefficient of performance
This is correct

For Part B and C
I am unsure how to approach this
please someone help me with an equation...

 

[Andrew Mason]

A Carnot refrigerator operating between -10.0 C and 40.0 C extracts heat from the cold reservoir at the rate 300 J/s. What are (a) the coefficient of performance of this refrigerator, (b) the rate at which work is done on the refrigerator and (c) the rate at which heat is exhausted to the hot side?...

For Part B and C
I am unsure how to approach this
please someone help me with an equation...

The co-efficient of performance of a refrigerator is the ratio of heat removed from the cold reservoir to the work done:

$$\text{COP} =\frac{Q_c}{W} = \frac{Q_c}{Q_h-Q_c} = \frac{Q_c/Q_h} {1 - Q_c/Q_h}$$

So W = Qc/COP. Since it is a Carnot cycle: Qc/Qh = Tc/Th

To find Qh use W = Qh-Qc

AM

 

[kyle8921]

I would approach this problem by using the equations and principles of thermodynamics. The Carnot refrigerator coefficient of performance is defined as the ratio of the desired output (cooling) to the required input (work). It can be calculated using the following equation:

Coefficient of performance (COP) = Q_c / W

Where Q_c is the heat extracted from the cold reservoir and W is the work done on the refrigerator.

(a) To calculate the coefficient of performance, we first need to convert the temperatures from Celsius to Kelvin. We get -10.0 C = 263.15 K and 40.0 C = 313.15 K. Plugging these values into the equation, we get:

COP = (263.15 K) / (313.15 K - 263.15 K)

COP = 0.84

So the coefficient of performance of this refrigerator is 0.84.

(b) To calculate the rate at which work is done on the refrigerator, we can use the following equation:

W = Q_h - Q_c

Where Q_h is the heat exhausted to the hot side and Q_c is the heat extracted from the cold reservoir. We know that Q_c is 300 J/s, so we just need to find Q_h. We can use the equation for the efficiency of a Carnot engine, which is the same as the efficiency of a Carnot refrigerator:

Efficiency = 1 - (T_c / T_h)

Where T_c and T_h are the temperatures in Kelvin. Plugging in the values, we get:

Efficiency = 1 - (263.15 K / 313.15 K)

Efficiency = 0.16

So the efficiency of the Carnot refrigerator is 0.16. Now we can use this to find Q_h:

Efficiency = Q_h / Q_c

0.16 = Q_h / 300 J/s

Q_h = 48 J/s

Therefore, the rate at which work is done on the refrigerator is 48 J/s.

(c) To calculate the rate at which heat is exhausted to the hot side, we can simply use the value we just found for Q_h, which is 48 J/s. This means that the rate at which heat is exhausted to the hot side is 48 J/s.

In conclusion, the coefficient of performance of this Carnot refrigerator is 0.84, the rate at