differential equations

The heating element in an electric kettle supplies heat to the kettle at a constant rate. This means that if there were no loss of heat from the kettle, the temperature of the water inside would rise at a constant rate. However,in practice the water loses heat at a rate which is proportional to the difference between the water temperature and the room temperature. At time t seconds the temperature of the water in the kettle is theta degrees celsius. The room temperature is assumed to be constant.

(i) Explain how the information given above leads to the differential equation
d(theta)/dt = H - K(theta-R),
where H,K and R are constant.

In a particular case, H = 1, K = 0.001 and R = 20.
(ii) show that in this case the differential equation in part (i) can be written in the form:
(1/1020-theta)d(theta/dt = 1/1000

(iii) Find the time taken for the water to heat up from 20 degrees to 100 degrees.

I have done (i) and (ii)

For (iii), I have separated the variables and integrated to give

- ln (1020 - theta) = 0,001t +c
I do not know how to proceed from here. I would be grateful for any advice or pointers. Thanks in advance.
(iii) Find the time

When you integrated, why not do the definite integral from (0,20) to (t,100) and solve for t as you've almost done.

Do you mean take the definite integral of both sides through the interval t to 0

( - ln( 1020 - theta)) in the interval t to 0 = (0.001t) in the interval t to 0?

Sorry if I am not making sense. I am confused.

Yes, from 20 to 100 (temperature) and from 0 to t (time).
Then solve for t.

Do you mean take the definite integral of both sides through the interval t to 0

( - ln( 1020 - theta)) in the interval t to 0 = (0.001t) in the interval t to 0?

Sorry if I am not making sense. I am confused.

The last part of the question
(iv) When the water temperature reaches 100 degrees the kettle switches off automatically, and no further heat is supplied. Find by how much the temperature of the water drops in the first 60 seconds after the kettle switches off.

d(theta)/dt = - 0,001(theta - 20)

ln(theta - 20) through the interval 100 to theta = - 0.001t through the interval 60 to 0

Am I on the right track? Again thanks for your time.

Temperature would go from 100 to theta.
Time would go from 0 to 60.

100 is the temperature at the initial time (0s).

Finished the question!! It was a complex one. Thank you