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 18 rubs, at a distance of 7.16 cm per rub, and with an average frictional force of 48 N, what is the temperature increase? The mass of tissues warmed is only 0.1 kg, mostly in the palms and fingers.

Answer 1

Answer:

0.178 °C

Explanation:

Given:

The frictional force, F = 48 N

Number of rubs, n = 18

Distance covered per rub, d = 7.16 cm = 0.0716 m

Total distance covered in 18 rubs, D = 0.0716 × 18 = 1.288 m

mass of the tissues warmed, m = 0.1 kg

Now,

Work done, W = Force × Displacement

W = 48 × 1.288 = 61.8624 J

Now,

Work done is the change in heat

W = ΔQ = 61.8624 J

also,

ΔQ = mCΔT

where, C is the specific heat

for human tissue, C = 3470 J/kg°C

thus,

61.8624 J = 0.1 × 3470 × ΔT

or

ΔT = 0.178 °C

Hence, the change in temperature is 0.178 °C