Heat Absorption by a Person's Hand from Steam to Liquid

[Lovely]

O.K.
HOw much heat is absorbed by a person's hand if steam at 100degree C is first converted to liquid water at 100 degree C and then cooled to normal body temperture?
I'm not sure of the formula for this one.

 

[russ_watters]

First, you need to know how much steam. Then, you multiply by the heat of vaporization to find how much energy is released in condensing it. Then, you take the specific heat, multiply by the mass and the temperature change to get the energy released in cooling it. Add the two numbers together.

 

[minger]

Set up your energy balance:
Q - W = U2 - U1
Q - W = m(h2 - h1)
Work here is 0, and we are looking for heat absorbed, or Q
Get out your steam tables, and look in the saturated water pressure tables, look at 100 kPa, and 100°C (likewise you can look in the steam tables). So, to first go from saturated vapor to saturated liquid, we go from Hg, to Hf. This difference in enthalpies is roughly 2260 kJ/kg (doing some quick mental math). Then, get to your saturated water temperature tables and find the Hf at whatever body temp is in SI units. Add these two numbers and that is your change in enthalpy. Multiply this by a mass to get your total heat absorbed by the hand.

Note that there is a LOT of energy associated by condensing water and a lot of energy needed to evaporate it.

Note that I also assumed that the enthalpy of condensed water is equal to the saturation enthalpy which is just a close approximation (I believe), but a good one.