A pot of water is heated to 100C and then placed in a room at a temperature of 18C. After 5 minutes, the pan temperature falls by 20C. Find the temperature after 10minutes.

(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.

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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.

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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

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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

and solve:

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)

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Seconly, we substitute in the boundary conditions.  Use T(t=0) = 100 to start with.

This gives:

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.

This gives:

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.

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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)