Question

If a woman rubs her hands back and forth for a total of 20 rubs, at a distance of 5.3 cm per rub, and with an average frictional force of 32.3 N, what is the temperature increase? The mass of tissues warmed is only 0.104 kg, mostly in the palms and fingers. Use a specific heat of 3500 J/kgºC.

Answer 1

Here, the work done by the friction force is equal to the amount of thermal energy which is transferred.

Work done W = Ff * Total distance in meters

5.3 cm = 0.053 m, Distance back and forth = 0.106 m
Total distance = 20 * 0.106 = 2.12 m

So, work done = 32.3 * 2.12 = 68.5 N * m

This is the amount of thermal energy which was transferred to the tissue.

Now -
?Q = mass * Specific heat * ?T
You need to know the specific heat of the tissue.

As given in the problem, the average specific heat of the human body is 3500 J/(kg * ?C)
So -

68.5 = 0.104 * 3500 * ?T

=> ?T = 68.5 / (0.104*3500) = 0.19 deg C.