(The rate at which the temperature of a body falls is proportional to the difference between the body temperature and the room temperature) - this would be given in the question.
This is a typical ordinary differential equation (ODE) question, which can be solved by separation of variables.
It is best looked at in three stages: Write the general ODE, then find the boundary conditons, then solve and find the equation for temperature in terms of time.
1. General ODE: let's call temperature T, and time t. Change is temperature over time is written as dT/dt. From the question, dT/dt is proportional to (T-18).
So dT/dt = -k(T-18) where k is the constant of proportionality and the minus sign arrises because the pan is cooling not heating.
2. Boundary conditions:
The pan of water is heated to 100C which is at t =0, so T(t=0) = 100.
After 5 minutes, the pan cools to 80C. So T(t=5) = 80
3. Solve the equation and substitute in the boundary conditions.
Firstly, we separate the variables and integrate.
dT/dt = -k(T-18)
dT = -k(T-18) dt
dT/(T-18) = -k dt
Add the integral signs (int):
int dT/(T-18) = int -k dt
ln(T-18) = -kt + C where C is the constant of integration.
Now do exp of both sides. exp is a way of wirting e to the power of ( ):
T - 18 = exp(-kt + C)
T - 18 = A exp(-kt) where A = exp(C). This is just used to simplify the problem.
T = 18 + A exp(-kt)
Seconly, we substitute in the boundary conditions. Use T(t=0) = 100 to start with.
100 = 18 + Aexp(0)
so A = 82
T = 18 + 82 exp(-kt)
Now it's time to use the 2nd boundary condition, T(t=5) = 80.
80 = 18 + 82 exp(-5k)
Subtract 18 from both sides, and divide by 82:
(80 - 18)/82 = exp(-5k)
Take ln of both sides and then divide by -5
k = -(1/5) ln(82/62)
k = 0.0559
So the final solution is:
T = 18 + 82 exp(-0.0559t)
We now have the general solution, which can be used to find any further information about the quesiton we need.
The question asks for the temperature at t=10. Simply substitute t-10 into the final solution above.
T = 18 + 82 exp(-0.0559 x 10)
T = 64.0C (three sig fig)