Question

How can i derive newton's law of cooling?

Answer 1

You can look up Wikipedia for its derivation.

Basically, the rate of cooling is proportional to the difference of the temperature between a hot object and its cooler environment. The more the difference the faster it cools in the beginning. As its temperature drops, the difference in temperature because smaller, therefore it cools slower in later time. So you expect it to cool rapidly in the beginning and much slower in the end. The is a typical exponential behavior.

Think of it this way, say a glass of boiling hot water is `100^@C` and the environment is at `0^@C` . The initial difference in temperature is `DeltaT`= `100^@C-0^@C=100^@C`.

Let the rate of change (in a given time interval) is 10% percent of the `DeltaT`, then the temperature of the water drops `10^@C to 90^@C`. The new `DeltaT= 90^@C`. So in the next time interval, the temperature drops another 10%, that is `9^@C`, to `81^@C`.

T ....................... Amount dropped (10%)
`100^@C` ....
`90^@C` ............. `10^@C`
`81^@C` .............. `9^@C`
`72.9^@C`............`8.1^@C`
`65.61^@C`......... `7.29^@C`
...
In the end, as `DeltaT` becomes smaller, the temperature drops much slower rate. This is a typical exponential decay (or negative compound) behavior. The math involves establishing a compound series. If you haven't learned calculus, just work out the infinite compound series.

Math (with calculus):

As explained above, the cooling rate is

`(dT)/dt ~~ - DeltaT`

Negative means the change in temperature is negative.

`DeltaT = T - T_E`
where T temperature of the object at time t, `T_E` is the temperature of the environment. Hence,

`DeltaT_0 = T_0 - T_E =` the difference in temperature initially,
where `T_0` is the initial temperature of the object.

Note that `dT = d(T- T_E) = d(DeltaT)`, because `dT_E`=0.

We can rewrite the equation above as:

`(dDeltaT)/dt = -rDeltaT`

where r is a cooling constant.

Rearrange to put T's on the same side and then integrate,

`int_(DeltaT_0)^(DeltaT)(dDeltaT)/(DeltaT)= -int_0^trdt`

`LnDeltaT|_(DeltaT_0)^(DeltaT) = -rt`

`Ln((DeltaT)/(DeltaT_0) )= -rt`

`DeltaT =DeltaT_0e^(-rt)`

Therefore, the final expression can be rewritten as:

`T = T_E+(T_0-T_E)e^(-rt)`

You can see how the temperature of the environmental play a role.

(Let me know if you need non-calculus derivation or you can look up the Web)