A freezer has a coefficient of performance of 6.30. It is advertised as using 39

Question

A freezer has a coefficient of performance of 6.30. It is advertised as using 396 kWh/yr. Note: One kilowatt-hour (kWh) is an amount of energy equal to running a 1-kW appliance for one hour. On average, how much energy does it use in a single day? On average, how much energy does it remove from the refrigerator in a single day? What maximum mass of water at 23.5 degree C could the freezer freeze in a single day? (The latent heat of fusion of water is 3.33 times 105 J/kg, and its specific heat is 4186 J/kg. degree C.)

Explanation / Answer

a) 396 kilowatt hours = 1425600000 joules So in single day it uses : 1425600000 joules / 365 = 3905753.425 J b) COP = Q2/ W So, Q2 = COP * W = 3905753.425 J* 6.3 = 24606246.58 J c) Q2 = M * (4186 * 23.5 + 333000) So, M = 57.042 kg

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.