Question

Q1: An electrical motor is used to operate a Carnot refrigerator with an interior temperature of 0°C. Liquid water at 0°C is placed into the refrigerator and transformed into ice at 0°C. If the room temperature is 300K, what mass of ice can be produced in one day by a 0.375-hp motor that is running continuously? Assume the refrigerator is perfectly insulated and operates at maximum theoretical efficiency

Answer 1

Answer:

7.5 x 10^5 g Ice

Explanation:

We know,

Hot T = 300 K

Cold T = 0°C = 0+273.15=273.15 K

Work done = 0.50 hp

Delta H of fusion of water is 6008 J per mol

Step 1

Convert power into J/s

1 hp = 746 W (J/s)

0.375 hp x 746 W/hp

0.375×746=279.75 J/s

Step 2

Calculate heat per unit time

Q = (cold T / hot T - cold T ) x power

(273.15/(300-273.15))×279.75=2914.71 J/s

Calculate moles of Ice

Moles of ice per second = Q/ delta H fusion

2914.71/6008=0.4851  mol /s

Step 4

Convert 1 day into second

1 day x ( 24 hours/day) x ( 3600 s / hour)

1×24×3600=86400 second

Step 5

Calculate moles of ice /day

0.4851 mol / s ) x 86400 s / day

0.6316*86400=41915.98 mol ice

Step 6

Use molar mass of ice and find mass in g

Molar mass is 18.0148 g/mol

Mass of ice  

41915.98×18.0148=755108.16061 g

Mass of ice produced per day is 755108.16061 g

Or 7.5 x 10^5 g

Answer is 7.5 x 10^5 g Ice