# 1.6 Mechanisms of heat transfer  (Page 11/27)

 Page 11 / 27

You can explore a simulation of the greenhouse effect that takes the point of view that the atmosphere scatters (redirects) infrared radiation rather than absorbing it and reradiating it. You may want to run the simulation first with no greenhouse gases in the atmosphere and then look at how adding greenhouse gases affects the infrared radiation from the Earth and the Earth’s temperature.

## Problem-solving strategy: effects of heat transfer

1. Examine the situation to determine what type of heat transfer is involved.
2. Identify the type(s) of heat transfer—conduction, convection, or radiation.
3. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.
4. Make a list of what is given or what can be inferred from the problem as stated (identify the knowns).
5. Solve the appropriate equation for the quantity to be determined (the unknown).
6. For conduction, use the equation $P=\frac{kA\text{Δ}T}{d}$ . [link] lists thermal conductivities. For convection, determine the amount of matter moved and the equation $Q=mc\text{Δ}T$ , along with $Q=m{L}_{\text{f}}$ or $Q=m{L}_{\text{V}}$ if a substance changes phase. For radiation, the equation ${P}_{\text{net}}=\sigma eA\left({T}_{2}{}^{4}-{T}_{1}{}^{4}\right)$ gives the net heat transfer rate.
7. Substitute the knowns along with their units into the appropriate equation and obtain numerical solutions complete with units.
8. Check the answer to see if it is reasonable. Does it make sense?

Check Your Understanding How much greater is the rate of heat radiation when a body is at the temperature $40\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ than when it is at the temperature $20\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ ?

The radiated heat is proportional to the fourth power of the absolute temperature . Because ${T}_{1}=293\phantom{\rule{0.2em}{0ex}}\text{K}$ and ${T}_{2}=313\phantom{\rule{0.2em}{0ex}}\text{K}$ , the rate of heat transfer increases by about 30% of the original rate.

## Summary

• Heat is transferred by three different methods: conduction, convection, and radiation.
• Heat conduction is the transfer of heat between two objects in direct contact with each other.
• The rate of heat transfer P (energy per unit time) is proportional to the temperature difference ${T}_{\text{h}}-{T}_{\text{c}}$ and the contact area A and inversely proportional to the distance d between the objects.
• Convection is heat transfer by the macroscopic movement of mass. Convection can be natural or forced, and generally transfers thermal energy faster than conduction. Convection that occurs along with a phase change can transfer energy from cold regions to warm ones.
• Radiation is heat transfer through the emission or absorption of electromagnetic waves.
• The rate of radiative heat transfer is proportional to the emissivity e . For a perfect blackbody, $e=1$ , whereas a perfectly white, clear, or reflective body has $e=0$ , with real objects having values of e between 1 and 0.
• The rate of heat transfer depends on the surface area and the fourth power of the absolute temperature:
$P=\sigma eA{T}^{4},$

where $\sigma =5.67\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}\phantom{\rule{0.2em}{0ex}}\text{J/s}·{\text{m}}^{2}·{\text{K}}^{4}$ is the Stefan-Boltzmann constant and e is the emissivity of the body. The net rate of heat transfer from an object by radiation is
$\frac{{Q}_{\text{net}}}{t}=\sigma eA\left({T}_{2}{}^{4}-{T}_{1}{}^{4}\right),$

where ${T}_{1}$ is the temperature of the object surrounded by an environment with uniform temperature ${T}_{2}$ and e is the emissivity of the object.

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