One further integration poser - due to popular demand

A pipe in a heating system contains hot water at a temperature of 80 degrees centigrade, and runs through a room where the background temperature is a constant 25 degrees centigrade.

The system is then drained.

Find the temperature of the pipe after 5 minutes, given that as it cools the temperature of the pipe obeys Newtons Law of Cooling in the form:

dT/dt = -0.5T

Where T is the difference between the pipe and the background, and t is the time in minutes.....

Reply 1

Fermat

T = 80
Ta = 25

dT/dt = -0.5(T - Ta)
dT/(T - Ta) = -0.5 dt

integrating both sides

ln(T - Ta) = -0.5t + C
at t = 0, T = 80, Ta = 25
ln(80 - 25) = C
C = ln55

Ln{(T - Ta)/55} = -t/2
T - Ta = 55.e^(-t/2)
================

at t = 5 mins

T - 25 = 55.e^(-2.5)
T - 25 = 4.5
T = 29.5
=======

Reply 2

Fermat

mattj35

... the temperature of the pipe obeys Newtons Law of Cooling in the form:

dT/dt = -0.5T

Where T is the difference between the pipe and the background, and t is the time in minutes.....

That should be,

dT/dt = -0.5ΔT

Where ΔT is the difference between the pipe and the background, and t is the time in minutes.....

And T is the temp of the cooling body.

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