Question

According to Newton’s law of cooling, the temperature of a body changes at a rate pro-

portional to the difference between its temperature and that of the surrounding medium (the ambient

temperature),

dT

dt = −k(T − Ta) with T(0) = T0

where T is the temperature of the body in oC, t is time in min, k = 0.019 min−1

is the proportion-

ality constant, and Ta = 20 oC is the ambient temperature. The analytical solution for this ordinary

differential equation (ODE) is:

T(t) = Ta + (T0 − Ta)e

−kt

The Euler’s numerical method for solving the ODE assumes that the first derivative is approximated

using the finite difference equation:

dT

dt ≈

∆T

∆t

=

Ti+1 − Ti

∆t

Suppose that a cup of coffee originally has a temperature of 70 oC. Solve the ODE

(a) Using Eulers method for the temperatures from t = 0 to 20 min using a step size of 5 min.

(b) Plot the results obtained in parts (a) together with the analytical solution given above

Answer 1

actually in this problem we are ananlysing the problem of body after every 5 min and applying Euler Equation

step size = 5 min =t

According to gven Euler equation here

dT/dt =T/t =    -k(T -Ta)

Initial Temperature of cup of tea = T_initial = 70 oC

Let T₁ ,T₂ ,T₃ ,T₄    be the temperature of body after first , 2nd, 3rd and 4th step size

given Ta =20 oC and k = 0.019 min^-1

after first step size of 5 min

T / t   = T / 5 = - k ( T₁ - Ta )      and     [    T₁   = T_initial - T    ]

T₁ = 65.66 oC , T ₂ = 61.69 oC , T₃ = 58.07 oC , T₄ = 54. 77 oC