Physics quiz question: chilling with a Whiskey Disk™

Physics 205A Quiz 7, fall semester 2012
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Practice Problem 14.5, Comprehensive Problem 14.75

"About Whiskey Disks™"
Whiskey Disks™/Rovettidesign

A 0.085 kg Whiskey Disk™ made of soapstone[*][**] at –10° C is placed into 0.22 kg of whiskey[***] at a temperature of 25° C. Ignore the effects of evaporation and phase changes, and heat exchanged with the environment and container. Specific heat of soapstone is 98 J/(kg·°C). Specific heat of whiskey is 3,400 J/(kg·°C). After reaching thermal equilibrium, the __________ had the greatest change in temperature.
(A) 0.085 kg Whiskey Disk™.
(B) 0.22 kg of whiskey.
(C) (There is a tie.)
(D) (Not enough information is given.)

[*] "Each disk weighs approximately 3 ounces," a.co/1TrbeVc.
[**] tulikivi.com/usa-can/fireplaces/Soapstone_characteristics.
[***] "About 3.4 J/(g·°C)," scottf.wordpress.com/2011/12/20/whiskey-stones-cooling-effectiveness/.

Correct answer (highlight to unhide): (A)

The transfer/energy balance equation is given by:

Qext = ∆Edisk + ∆Ewhiskey,

Qext = mdisk·cdisk·∆Tdisk + mwhiskey·cwhiskey·∆Twhiskey.

Since there are no heat exchanges with the environment or the container, the Whiskey Disk™ and the whiskey only exchange heat with each other, such that:

0 = mdisk·cdisk·∆Tdisk + mwhiskey·cwhiskey·∆Twhiskey,

mdisk·cdisk·∆Tdisk = mwhiskey·cwhiskey·∆Twhiskey,

Tdisk/∆Twhiskey = –mwhiskey·cwhiskey/mdisk·cdisk,

Tdisk/∆Twhiskey = –((0.22 kg)·(3,400 J/(kg·°C)))/((0.085 kg)·(98 J/(kg·°C))),

Tdisk = –89.79591837·∆Twhiskey,

(or to two significant figures, this numerical factor is –90), which means that the ∆Tdisk increase in temperature is ninety times greater than the ∆Twhiskey decrease in temperature.

Sections 70854, 70855
Exam code: quiz07Di5k
(A) : 24 students
(B) : 10 students
(C) : 16 students
(D) : 0 students

Success level: 48%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.78

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