Question

Rubbing your hands together warms them by converting work into thermal energy. If a woman rubs her hands back and forth for a total of 20 rubs, at a distance of 7.50 cm per rub, and with an average frictional force of 40.0 N, what is the temperature increase? The mass of tissues warmed is only 0.100 kg, mostly in the palms and fingers.

(a) 3.82°C
(b) 5.67°C
(c) 2.45°C
(d) 6.91°C

Answer 1

The temperature increase due to friction when the woman rubs her hands together is 1.43°C. The given options don't match this result, suggesting there might be missing information or an error in the options.

Step-by-step explanation:

Calculating the Temperature Increase from Friction

When rubbing hands together, the work done against friction is converted into thermal energy, warming up the hands. To calculate the temperature increase, we'll use the formula for work: Work = Force × Distance × Number of Rubs. With a frictional force of 40.0 N, a distance of 7.50 cm per rub, and 20 rubs, the total work done is calculated in joules (J). Then, we apply the formula Q = mcΔT where Q is the heat gained, m is the mass, c is the specific heat capacity of human tissue (assumed to be the same as water, 4.186 J/g°C, since the body is mostly water), and ΔT is the temperature change.

First, we convert the distance per rub to meters (0.075 m) and calculate the total work done:

• Work = 40.0 N × 0.075 m × 20 = 60.0 J

Next, we determine the temperature increase using the formula for heat transfer:

• mcΔT = 60.0 J
• (0.100 kg × 1000 g/kg) × 4.186 J/g°C × ΔT = 60.0 J
• ΔT = 60.0 J / (0.100 kg × 1000 g/kg × 4.186 J/g°C)
• ΔT = 1.43°C

The temperature increase of the tissues in the woman's hands due to the friction during rubbing is 1.43°C, which is not one of the provided options. There is likely an error in the given options or an additional piece of information required that has not been supplied.