Heat transfer via radiation is something we've spoken about already in the quantization and waves chapter. It is black body radiation, or light emitted due to the temperature of an object. This is sometimes also called thermal radiation. The power emitted by an object at temperature T is [math]P=\epsilon\sigma AT^4[/math]. Recall that the Stefan Boltzmann constant [math]\sigma = 5.67\times10^{-8} \tfrac{W}{m^2\cdot K^4}.[/math] The power radiated can also be seen as heat[i] lost[/i] per unit time by the system, or [math]P=\frac{-dQ}{dt}[/math]. Recall that when we discussed the effectiveness of a space heater (or a human acting as one) that the heater also received heat from the environment. So the rate of heat transfer depends not only on the object's temperature, but also on the environmental temperature since the environment radiates energy toward the object. In this case [math]P=\epsilon\sigma A\left(T^4-T_{environment}^4\right)[/math]. So the final equation for heat transfer via radiation is[math]\frac{dQ}{dt}=-\epsilon\sigma A\left(T^4-T_{environment}^4\right)[/math]. In the same way as with conduction, we can relate this to the rate of change of temperature if we know the mass and specific heat capacity of the object that's radiating. So we can write:[center] [math]\frac{dT}{dt}=-\frac{\epsilon\sigma A}{mc}\left(T^4-T_{environment}^4\right).[/math][/center]The plot of temperature versus time for this equation is generally arrived at via numerical methods. In limited cases, only in the past few years have solutions to such differential equations been arrived at analytically as in the paper below.

EXAMPLE: Suppose we have a red-hot (T=900K) cast iron ball of radius 10cm. The material has an emissivity of 0.70. How much power does this ball radiate to the environment?[br]ANSWER: [math]P=\epsilon\sigma AT^4[/math] and [math]A=4\pi r^2.[/math] Plugging in gives P=3272W.[br][br]EXAMPLE: If the ball sits in a room held at 25C, at what rate is energy absorbed by the ball from the room?[br]ANSWER: Rate of energy absorption is also just power. So [math]P=\epsilon\sigma AT_{room}^4=39W.[/math][br][br]EXAMPLE: At the current temperature, at what rate is the temperature of the ball decreasing? Based on your values, approximate the time it will take for the ball to cool to 800K.[br]ANSWER: We need to find the amount of heat loss (Q) needed to reduce the ball's temperature by 100 degrees C. [math]Q=mc\Delta T.[/math] To find this we need the mass of the ball. The density of iron is [math]\rho=7.87\times10^3kg/m^3.[/math] So the mass is [math]m=\rho V=7.87\times10^3kg/m^3\left(\frac{4}{3}\pi r^3\right)=33kg.[/math][br]The rate of heat loss depends on temperature (which lowers as heat is lost). But if we use the average temperature (decent estimate) of 850K, we can get a rate of heat loss of [math]P=0.70(5.67\times 10^8 W/m^2/K^4)4\pi (0.1m)^2 850K=2565W[/math] when accounting for the 39W coming back from the environment. Taking the time derivative of the heat equation above, we get [math]\frac{dQ}{dt}=mc\frac{dT}{dt}[/math] and the rate of change of heat is just power P (with a minus sign). So [math]-P=-2565W = mc\frac{dT}{dt}.[/math] Last, we solve for [math]\frac{dT}{dt}[/math] and get 0.173 degrees C per second as the cooling rate. At this (assumed average) rate, it would take 579s, or a little under ten minutes to cool by 100 degrees from 900K to 800K.