Question

A certain refrigerator has a COP of 5.00. When the refrigerator is running, its power input is 500 W. A sample of water of mass 600 g and temperature 20.0°C is placed in the freezer compartment. How long does it take to freeze the water to ice at 0°C? Assume all other parts of the refrigerator stay at the same temperature and there is no leakage of energy from the exterior, so the operation of the refrigerator results only in energy being extracted from the water.

Answer 1

Answer:

t=20s

Explanation:

To solve this problem we must apply the first law of thermodynamics, which indicates that the energy that enters a system is the same that must come out, resulting in the following equation

For this problem we will assume that the water is in a liquid state, since it is a domestic refrigerator

q=m.cp.(T2-T1)

q=heat

m=mass of water =600g=0.6Kg

cp=

specific heat of water=4186J/kgK

T2=temperature in state  2=20°C

T1=temperature in state 1=0°C

solving:

q=(0.6)(4186)(20-0)=50232J

A refrigerator is a device that allows heat to be removed to an enclosure (Qin), by means of the input of an electrical energy (W) and the heat output (Qout), the coefficient of performance COP, allows to know the ratio between the heat removed ( Qin) and the added electrical power (W), the equation for the COP is

$COP=\frac{Qin}{Win}$

To solve this exercise we must know the value of the heat removed to the water (Qin)

solving for Qin

Qin=(COP)(Win)

Qin=(5)(500W)=2500W

finally we remember that the definition of power is the ratio of work over time

w=work

p=power=500w

$P=\frac{W}{t} \\t=\frac{W}{P} \\t=\frac{50232}{2500} \\t=20.09s$