A modern version of Joule's experiment has you turn a crank attached to an 85.0 g aluminum cylinder. A belt exerts a frictional torque on the cylinder of 1.50 Nm. The specific heat of aluminum is 900 J/(kg·K). How many turns of the crank are necessary to raise the temperature of the aluminum by 40 C°?
A) 325
B) 650
C) 900
D) 975
E) 1300

Question

Quizplus · Accepted Answer

The work done by the torque is equal to the change in thermal energy of the aluminum. The number of turns of the crank is equal to the work done divided by the torque.$$W = 	au 	heta = (1.50 	ext{ Nm}) 	heta$$$$\Delta U = mc_p \Delta T = (0.085 	ext{ kg})(900 	ext{ J/kg$\cdot$K})(40 	ext{ K}) = 3060 	ext{ J}$$$$	heta = \frac{\Delta U}{W} = \frac{3060 	ext{ J}}{1.50 	ext{ Nm}} = 2040 	ext{ rad}$$$$N = \frac{	heta}{2\pi} = \frac{2040 	ext{ rad}}{2\pi} = 325 	ext{ turns}$$

A Modern Version of Joule's Experiment Has You Turn a Crank