## 20111215

### Physics quiz question: net heat rate radiated by stove

Physics 205A Quiz 7, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 14.65

A black wood stove has a surface area of 1.6 m2 and a surface temperature of 180° C. The room temperature is 15° C. What is the net rate at which heat is radiated into the room?
(A) 67 watts.
(B) 95 watts.
(C) 3.2x103￼ watts.
(D) 3.8￼x103 watts.

The rate at which the wood stove radiates heat to the environment is given by:

(heat flow)/time = e*σ*A*(Tobj)4 = 3,820 watts,

which is response (D), where the temperature of the object is given in kelvins: 180° C + 273 = 453 K.

The rate at which the wood stove absorbs heat from the environment is given by:

(heat flow)/time = e*σ*A*(Tenv)4 = 624 watts,

where the temperature of the environment is given in kelvins: 15° C + 273 = 288 K.

Thus the net rate at which heat is radiated into the environment is then:

(heat flow)/time = e*σ*A*[(Tobj)4 - (Tenv)4] = 3,196 watts,

or to two significant figures, 3.2x103 watts.

Response (A) uses the temperature difference raised to the fourth power (Tobj - Tenv)4; response (B) is the power radiated by the stove, but using 180° C instead of 453 K.

Sections 70854, 70855
Exam code: quiz07h3A7
(A) : 3 students
(B) : 24 students
(C) : 23 students
(D) : 2 students

Success level: 44%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.46