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### Heat Supplied and Heat Loss of a Furnace

```
Date: 05/17/2000 at 21:08:02
From: Peter Lehman
Subject: Differential Equations

A furnace is switched on at 9 a.m. Heat is supplied at a constant
rate, but as the furnace increases in temperature, heat is lost at a
rate determined by the difference between its temperature and the
temperature of its surroundings. The temperature of both the furnace
and its surroundings was 20 C at 9:00 a.m. At 9.30 a.m the furnace was
200 C and by 10:00 a.m it was 350 C. The minimum operating temperature
is 800 C. Use a differential equation to find the time when the
furnace is ready for use, and the highest temperature it could reach.

It is just the differential equation I need help on. The rest is okay.
```

```
Date: 05/18/2000 at 08:16:28
From: Doctor Anthony
Subject: Re: Differential Equations

Suppose that heat is added at H units per min. It will be lost at a
rate proportional to (T-20).

The increase in heat per unit time, dh/dt will be given by:

dh/dt = H - k(T-20)

where k is a constant of proportionality.

For a constant mass and specific heat we can say that dT/dt is
proportional to dh/dt and rewrite the differential equation in the
form:

dT/dt = a - bT

dT/(a-bT) = dt

-(1/b)ln(a-bT) = t + C

ln(a-bT) = -bt + C'

a-bT = A.e^(-bt)

bT = a - A.e^(-bt)

T = (1/b)[a - A.e^(-bt)]

at t = 0, T = 20 and so:

20 = (1/b)[a - A]

20b = a - A  so  a = A+20b

at t = 30, T = 200:

200 = (1/b)[A+20b - A.e^(-30b)]

200b = A + 20b - A.e^(-30b)

180b = A[1 - e^(-30b)]

A = 180b/[1-e^(-30b)]

at t = 60, T = 350:

350 = (1/b)[a - A.e^(-60b)]

350b = a - A.e^(-60b)

350b = A + 20b - A.e^(-60b)

330b = 180b[1 - e^(-60b)]/[1-e^(-30b)]

330[1-e^(-30b)] = 180[1-e^(-60b)]

11 - 11e^(-30b) = 6 - 6e^(-60b)

6e^(-60b) - 11e^(-30b) + 5 = 0

putting e^(-30b) = u, this equation is a quadratic in u.

6u^2 - 11u + 5 = 0

(6u-5)(u-1) = 0

so u = 1 or u = 5/6

If u = 1, e^(-30b) = 1  and b = 0  (we can ignore this solution)

so u = 5/6,  e^(-30b) = 5/6

-30b = ln(5/6)

b = 0.0061

Then

A = 180b/[1-e^(-30b)] = 180b/[1-5/6] = 180b/(1/6) = 1080b

= 6.5636

and

a = A+20b = 6.5636 + 0.1215 = 6.68515

Finally we have:

T = (1/b)[a - A.e^(-bt)]

=  164.5[6.68515 - 6.5636e^(-0.0061t)]

=  1100.00 - 1080 e^(-0.0061t)

Clearly the highest possible temperature will be 1100 degrees.

Time to 800 degrees is given by:

800 = 1100 - 1080 e^(-0.0061t)

1080 e^(-0.0061t) = 300

e^(-0.0061t) = 5/18

-0.0061t = ln(5/18)

t = 209.99 minutes

And so time to 800 degrees = 210 minutes.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus
High School Physics/Chemistry

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