Answer to Question #117522 in Molecular Physics | Thermodynamics for HH

Question

You are jogging along at a slow speed on a cool day and are losing heat at a rate of 120 W via convection. If your surface area is 0.45 m2, your skin temperature is 26°C. and the air temperature is 17°C then what is the heat transfer coefficient of convection, hconv? (to 2 s.f and in W m−2 K−1)

Answer 1

By definition, the heat transfer coefficient of convection is given by the expression (see https://en.wikipedia.org/wiki/Heat_transfer_coefficient):

$$h_{conv} = \dfrac{q}{\Delta T},$$

where $q = P/A$ is the heat flux (thermal power per unit area) and $\Delta T$ is the difference in temperature (in Kelvin) between the body surface and air. Putting it all together:

$$h_{conv} = \dfrac{P}{A\Delta T} =\dfrac{120W}{0.45m^2((273+26)K-(273+17)K)} \approx29.6 \dfrac{W}{m^2K}$$

Answer. hconv = 29.6 W m-2 K-1.