A refrigerator door is opened and room-temperature air (20.0 degreeC) fills the

Question

A refrigerator door is opened and room-temperature air (20.0 degreeC) fills the 1.50 m^3 compartment. A 15.0 kg turkey, also at room temperature, is placed in the refrigerator and the door is closed. The density of air is 1.20 kg/m^3 and its specific heat is 1020J/(kg middot K). Assume the specific heat of a turkey, like that of a human, is 3480 J/(kg middot K). How much heat must the refrigerator remove from its compartment to bring the air and the turkey to thermal equilibrium at a temperature of 5.00 degreeC ? Assume no heat exchange with the surrounding environment.

Explanation / Answer

Find the heat removed from the air:
heat = mass*specific heat*difference in temperature
mass of 1.50 m^3 air = 1.50m^3*1.20kg/m^3 = 1.80kg
heat removed = 1.80kg*1020J/(kg*K)*15K = 27540 J

Find heat removed from the turkey:
heat removed = 15.0kg*3480J/(kg*K)*15K = 783000 J

total heat = heat from turkey + heat from air = 27540J +783000J = 810540 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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