# Time for black body to cool to a given temperature

I'm trying to figure out the time required for a blackbody to cool assuming it only looses heat via radiation.

I can estimate the mass, specific heat, surface area, emissivity, initial temperature, final temperature, etc.

I know the amount of heat which has to be lost is:

$$Q = mc\Delta T$$,

where $$m$$ is mass, $$c$$ is specific heat, $$\Delta T$$ is a temperature_change

I know the rate of heat loss is:

$$P = \epsilon kST^4$$,

where $$k$$ is Stefan Boltzmann, $$T$$ is the body temperature, $$S$$ is the surface area, $$\epsilon$$ is the emissivity.

The body temperature changes as the blackbody cools so I cannot simply use the initial or final temperature or I get very different results. I think I have to somehow integrate $$P$$ from $$t = 0$$ (when body is at initial temperature) to when the body is at the final temperature.

Any help would be greatly appreciated.

• Is the final temperature the same as your surroundings? – JMac Jan 15 at 15:22

Let us assume temperature of black body is $$T_1$$ and of surrounding is $$T_2$$, also $$T_1>T_2$$ , at time $$t =0$$. Heat lost by body at any instant is

$$F= k*S_{area}*(T^4 -T_2^4)$$.

Where T is temp at that instant and $$k$$ is the Boltzmann constant and $$S_{area}$$ is the surface of the body. Now this lost heat can be written as

$$F= - ms \frac{dT}{dt}$$

• dT : small change in temp
• dt : small change in time

$$-ms \frac{dT}{dt} = \sigma(T^4 - T_2^4 )S_{area}$$

Integrate this expression and put the limits and solve.

• Thanks @Jaskeerat. I got about that far myself but my integration is a bit rusty. Luckily I discovered online integration calculators can help me, e.g.: [link](wolframalpha.com/input/?i=1%2F(x%5E4-a%5E4). Thanks very much for your contribution. – Marcus Jan 17 at 0:16

Two things you need to remember:

1. The body is absorbing radiation from its surroundings at all times. If you don't account for this effect, your math will suggest that the body continues to cool until it reaches absolute zero. If you account for this effect properly, the final temperature of the body will be equal to the temperature of the surroundings.

2. You will need to account for the fact that thermal energy in the core of the body needs to be convected/conducted to the surface before it can be radiated away. The easiest way to account for this would be to assume that the convection/conduction is very fast and thus that the body effectively has a uniform temperature, but this approximation becomes less appropriate as the surface area-to-volume ratio gets smaller and as the body becomes worse at convecting/conducting heat.

If you can relate the rate of heat loss to the current temperature of the body, then you can generate a differential equation for internal energy (or, using specific heats, for temperature). You should be able to solve this equation for internal energy or temperature as a function of time. Note that the time to cool to the surroundings temperature will always be infinity.

• Thanks @user1476176. I omitted the surrounding temperature dependence of the heat loss rate to simplify my question. Your second point about the temperature gradient inside the black body can also be ignored in my case as I'm dealing with tiny particles of a very good conductor. Thanks for the input though. – Marcus Jan 17 at 2:17