The inside of an ideal refrigerator is at a temperature Tc, while the heating co

Question

The inside of an ideal refrigerator is at a temperature Tc, while the heating coils on the back of the refrigerator are at a temperature Tb. Owing to a malfunctioning switch, the light bulb within the refrigerator remains on when the door is closed. The power of the light bulb is P; assume that all of the energy generated by the light bulb goes into heating the inside of the refrigerator. For all parts of this problem, you must assume that the refrigerator operates as an ideal Carnot engine in reverse between the respective temperatures. If the temperatures inside and outside of the refrigerator do not change, how much extra power Pextra does the refrigerator consume as a result of the malfunction of the switch? Express the extra power in terms of P, Th, and Tc. Suppose the refrigerator has a 25-W light bulb, the temperature inside the refrigerator is 0 degree C, and the temperature of the heat dissipation coils on the back of the refrigerator is 35 degree C Find the extra power Pextra consumed by the refrigerator. Keep in mind that you will need to use absolute units of temperature (i.e., kelvins). Express your answer numerically in watts to three significant figures.

Explanation / Answer

A) COP = Tc / (Th - Tc) = P / Pextra

Pextra = P / [Tc / (Th - Tc)]

B) Tc = 0 C = 273 K

Th = 35 C = 308 K

P = 25 W

Using Pextra = P / [Tc / (Th - Tc)]

Pextra = 25 / [273 / (308 - 273)]

   = 3.205 W

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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