Using Laplace transformation please help me solve the following?

Question

The Big Pot of Soup As part of his summer job at a resturant, Jim learned to cook up a big pot of soup late at night, just before closing time, so that there would be plenty of soup to feed customers the next day. He also found out that, while refrigeration was essential to preserve the soup overnight, the soup was too hot to be put directly into the fridge when it was ready. (The soup had just boiled at 100 degrees C, and the fridge was not powerful enough to accomodate a big pot of soup if it was any warmer than 20 degrees C). Jim discovered that by cooling the pot in a sink full of cold water, (kept running, so that its temperature was roughly constant at 5 degrees C) and stirring occasionally, he could bring the temperature of the soup to 60 degrees C in ten minutes.
How long before closing time should the soup be ready so that Jim could put it in the fridge and leave on time ?

Answer 1

The formula for heat loss is

dT/dt = -k(T-Tw) (Tw = 5ºC)

Newton's law of cooling, see http://en.wikipedia.org/wiki/Heat_conduction

Re-write this as

dT/dt + kT = kTw

Using Laplace transforms, sT(s) + kT(s) -T(0) = kTw/s

T(0) = 100ºC

sT(s)+kT(s) = T(0) + kTw/s

T(s)*(s + k) = T(0) + kTw/s

T(s) = T(0)/(s + k) + kTw/[s(s + k)]

The inverse transform is

T(t) = T(0) e^-kt+ Tw(1 - e^-kt)

http://en.wikipedia.org/wiki/Laplace_transform

This can be written T(t) = Tw - (Tw-T0)e^-kt

T(10) = 60, solve for k.

Then solve for t such that T(t) = 20ºC

Answer 2

First you find the Laplace transform your functions. Solve for Y(s), then use inverse Laplace transform table to find your solution. example: y" + y = - 2 sin (t) s^2 Y(s) -sY(0) -Y(0) + Y(s)= -2/(s^2+1) (s^2+1)Y(s) - (-s-1)Y(0) = -2/(s^2+1) Y(s) = (-2/(s^2+1) + (-s-1)Y(0))/(s^2+1) Plug in your initial conditions, simplify and use an inverse Laplace transform table (or the convolution integral)