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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 11 rubs a distance of 7.50 cm each and with a frictional force averaging 71.3 N, what is the temperature increase? The mass of tissue warmed is only 0.100 kg, mostly in the palms and fingers. The specific heat of the tissue is 3500 J/(kg · °C).

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Answer:

0.168°C

Explanation:

The frictional force between the palms enables rubbing and initiates the heat deposed.

The energy generated per rub is;

The force × distance

Hence we have; 71.3 N×0.0750m

Hence for 11 rubs we have; 11× 71.3 N×0.0750m= 58.82J

Now this heat acquired by the tissue is the same as mass of tissue × specific heat capacity of tissue × temperature rise;

Expressed mathematically as;

58.82=0.100 kg×3500 J/(kg · °C)×∆T

∆T= 58.82/ 0.100 ×3500 =0.168°C

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