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) 3.8x103 watts.
Correct answer: (C)
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
20111215
Physics quiz question: net heat rate radiated by stove
Labels:
blackbody,
emissivity,
heat transfer,
radiation,
Stefan-Boltzmann law,
temperature
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