# Question Video: Heat Transfer

Rubbing hands together warms them by converting work into heat. A woman rubs her hands back and forth for a total of 35 rubs, where the hands move a distance of 9.0 cm across each other with each rub, against an average frictional force of 31 N. Determine the temperature increase of the warmed parts of her hands, which have a mass of 0.080 kg and a specific heat capacity of 3500 J/kg⋅°C. Assume no heat loss from the hands during the rubbing process.

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### Video Transcript

Rubbing hands together warms them by converting work into heat. A woman rubs her hands back and forth for a total of 35 rubs, where the hands move a distance of 9.0 centimeters across each other with each rub, against an average frictional force of 31 newtons. Determine the temperature increase of the warmed parts of her hands, which have a mass of 0.080 kilograms and a specific heat capacity of 3500 joules per kilogram degrees Celsius. Assume no heat loss from the hands during the rubbing process.

In this scenario, we have a woman rubbing her hands back and forth together to heat them up. We’re told that each rub covers a distance of 9.0 centimeters and works against a frictional force of 31 newtons. We want to solve for the increase in temperature, Δ𝑇, of her hands after she rubs them together 35 times.

To begin solving for Δ𝑇, we can first calculate the work done by the woman as she rubs her hands together. Work, we recall, is equal to force times displacement. This means that the total work done by the woman is equal to the force she opposes, the frictional force, times the distance over which she opposes that force times the number of times she rubs her hands together. Or we could think of 𝑑 times 𝑛 as the total distance that we use in this calculation. Either way, her total work is equal to 𝐹 sub 𝑓 times 𝑑 times 𝑛.

We plug in 31 newtons for 𝐹 sub 𝑓 and the displacement in units of meters for 𝑑 and 35 for the number of times she rubs her hands together. As we consider the units of this expression, newtons times meters, we realize that a newton times a meter is a joule, a unit of energy.

We can write the total work done by the woman therefore as 31 times 0.09 times 35 joules. It’s this energy that is converted into heat through friction, which increases the temperature of the woman’s hands. This temperature increase, Δ𝑇, is based on the specific heat capacity of the woman’s hands as well as their mass.

Considering the specific heat capacity, let’s look for a moment at the units of this term. Those units are joules, units of energy, divided by kilograms, units of mass, times degrees Celsius, units of temperature. We want to solve ultimately for a change in temperature, which will have units of degrees Celsius. This means that, en route to getting there, we’ll want to invert the units of the specific heat capacity; that is, we’ll want to use one over 𝐶 sub 𝑃.

If we take this inverse, then multiply it by the total work done by the woman, and divide that by the mass of the woman’s hands, then from a units perspective, we’ll see the units of joules cancel out, as well as the units of kilograms, leaving us with units of degrees Celsius, just what we want for Δ𝑇.

We can say then that Δ𝑇 is equal to the total work done by the woman divided by the mass of the hands that are being heated up multiplied by the specific heat capacity of those hands. Plugging in for these values and entering the expression on our calculator, we find Δ𝑇 is 0.35 degrees Celsius. That’s how much the woman’s hands warm up by rubbing them together this way.