Minimum amount of work input to a refrigerator

[8614smith]

Homework Statement

600Kg of water at 25C has to be frozen into ice in an ideal refrigerator. The room temperature is 20C. What is the minimum amount of work input to the refrigerator to achieve this? ( specific heat capacity of water = 4170 J/Kg/K, latent heat of melting ice = 3.33 x10^5 J/Kg

Homework Equations

$${Q_C}=mL$$

$${Q_C}=mc\delta{T}$$

$$C.O.P=\frac{Q_C}{W}=\frac{Q_C}{{Q_H}-{Q_C}}=\frac{{T_C}/{T_H}}{1-{T_C}/{T_H}}$$

The Attempt at a Solution

The refrigerator has to remove heat equal to 600Kg x c x dT to get the temperature down to 273, then has to remove heat from a 273K reservoir to a 293K reservoir to keep it frozen.

1. cooling to 273, heat removed is $${Q_C}=mc\delta{T}$$ = $${Q_C}=600*4170*25=6.255E^{6}Joules$$

2. keeping it frozen - amount of heat that needs to be removed is $${Q_C}=mL=600*3.33E^{5}=1.998E^8 Joules$$

so total is the sum of these = 2.06E^8 J

then using $$C.O.P=\frac{Q_C}{W}=\frac{Q_C}{{Q_H}-{Q_C}}=\frac{{T_C}/{T_H}}{1-{T_C}/{T_H}}$$

i get two different coefficients of power - one for cooling and one for freezing, what do i do with them, do i add them or subtract one from the other? or do i use the sum of the two energies and put this into the COP? or am i doing this completely wrong?

 

[mark726]

Thank you for your question. To calculate the minimum amount of work required for the ideal refrigerator to freeze 600kg of water at 25C into ice at 0C, we need to consider the two steps involved in this process.

Step 1: Cooling the water from 25C to 0C
In this step, the refrigerator needs to remove heat from the water to bring its temperature down to 0C. The amount of heat removed can be calculated using the formula {Q_C}=mc\delta{T}, where m is the mass of water (600kg), c is the specific heat capacity of water (4170 J/Kg/K), and dT is the change in temperature (25C-0C=25K). Substituting these values, we get {Q_C}=600*4170*25=6.255E^{6}Joules.

Step 2: Freezing the water at 0C into ice at 0C
In this step, the refrigerator needs to remove heat from the water at 0C to freeze it into ice at 0C. The amount of heat removed can be calculated using the formula {Q_C}=mL, where m is the mass of water (600kg) and L is the latent heat of melting ice (3.33 x10^5 J/Kg). Substituting these values, we get {Q_C}=600*3.33E^{5}=1.998E^8 Joules.

Total amount of heat removed = 6.255E^{6}Joules + 1.998E^8 Joules = 2.06E^8 J

To calculate the minimum amount of work input to the refrigerator, we need to consider the Coefficient of Performance (COP). The COP is the ratio of the heat removed from the cold reservoir (water) to the work input to the refrigerator. It can be calculated using the formula C.O.P=\frac{Q_C}{W}=\frac{Q_C}{{Q_H}-{Q_C}}=\frac{{T_C}/{T_H}}{1-{T_C}/{T_H}}, where Q_C is the heat removed from the cold reservoir, Q_H is the heat supplied to the hot reservoir (room temperature), T_C is the temperature of the cold reservoir (0C), and T_H is the temperature of the hot reservoir (20C