Question

I'd need help with the whole question..but even just in part 9i, the answer is d(theta)/d(t)=kt. Where does the t come from?? It's simple things like this which confuse me.. Thanks for your help!!

Answer 1

(i) `t = 20/k_1 \ s`

(ii) `(d theta)/dt = - k_2(theta-20)`

(iii) `t = (ln2)/k_2 + 20/k_1 \ s`

Explanation:

Here `t` represents the time (in seconds) since the heater is switched on.

We are given that `theta=40` when the heater is first switched on, ie when `t=0`

Also, that the temperature increases at a constant rate of `k_1` degrees Celsius per second, ie:

`(d theta)/dt = k_1 \ \ \ \` where `k_1` is a constant

Part (i):

This is a First Order Separable (and trivial) Differential Equation,. If we directly integrate we get:

`theta = k_1t + C`

Given that `theta=40` when `t=0`, then

`40 = 0 + C => theta = k_1t + 40`

And so when `theta =60` then:

`60 = k_1t + 40 => k_1t=20 => t = 20/k_1`

Part (ii):

We now have that the rate of change of temperature wrt `t` decreases at a variable rate of `k_2(theta-20)`degrees Celsius per second, ie:

`(d theta)/dt = - k_2(theta-20)` ..... [A]

Where for the cooling process we have `theta=60` when `t=0`

Part (iii):

Denote the total time sought for the temperature to decrease to `theta=40` by `T`,

The DE [A] is a First Order Separable DE, so we can separate the variables:

`int 1/(theta-20) \ d theta = - k_2 \ int \ dt`

Which we can nioe integrate to get:

`ln|theta-20| = - k_2t + C`

We can use the initial condition `theta=60` when `t=0` we have:

`ln(60-20) = 0 + C => C = ln 40`

Noting that `theta gt 20` over the given range, we then have:

`ln(theta-20) = - k_2t + ln40`

So when `theta=40 => ln(40-20) = - k_2T + ln40`

`:. k_2T = ln40-ln20 => T = ln(40/20)/k_2`

Hence, the total time sought is given by:

`t = ln(2)/k_2 + 20/k_1`