Answer to Question #122480 in Differential Equations for jse

Question

1. A thermometer reading 20° C is placed in an oven preheated to a constant temperature. Through a glass window in the oven door, an observer records that the thermometer reads 42° C after
1/2 minute and 62° C after 1 minute. How hot is the oven?

_____ degrees Celsius

Answer 1

Let T represents temperature of thermometer and S be the constant temperature of oven.

t represents time

Here all temperatures are in °C and time in minute

$\frac {dT}{dt}$ is proportional to differenc of temperature of thermometer and that of oven.

So $\frac {dT}{dt}$ = k(T-S), k is constant

=> $\frac {dT}{T-S}$ = k dt

Integrating

$\int$ $\frac {dT}{T-S}$ = $\int$ k dt

ln | T-S | = kt + C , C is integration constant

=> ln(S-T) = kt + C as S > T here

When t = 0 , T = 20

So ln(S-20) = k*0 + C

=> C = ln(S-20)

So ln(S-T) = kt + ln(S-20)

=> ln(S-T) - ln(S-20) = kt

=> ln $\frac {S-T}{S-20}$ = kt

Two set of values of t, T are as follows

When t = $\frac {1}{2}$ , T = 42

and when t = 1, T = 62

So ln $\frac {S-42}{S-20}$ = $\frac {k}{2}$ ..........eq(1)

and ln $\frac {S-62}{S-20}$ = k ............eq(2)

Comparing eq(1) and eq(2)

2 ln $\frac {S-42}{S-20}$ = ln $\frac {S-62}{S-20}$

=> ln $[\frac {S-42}{S-20} ]^2$ = ln $\frac {S-62}{S-20}$

=> $[\frac {S-42}{S-20} ]^2$ = $\frac {S-62}{S-20}$

=> $\frac {(S-42)^2}{(S-20)}$ = (S-62) as S>20, (S - 20)≠0

=> (S-42)² = (S-20)(S-62)

=> S² - 84S + 1764 = S² - 82S + 1240

=> - 84S + 1764 = - 82S + 1240

=> -84S + 82S = 1240 - 1764

=> -2S = -524

=> S = 524/2 = 262

The oven is 262 degree Celsius