Question

A thermometer reading 65° F 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 110° F after 1 2 minute and 140° F after 1 minute. How hot is the oven?

Answer 1

The temperature of the oven is 200°F

Explanation:

Given-

We have to apply Newton's law of cooling or heating

$\frac{dT}{dt} = k ( T - Tm)\\\\\frac{dT}{T - Tm} = kdt$

On Integrating both sides, we get

$T = Tm + Ce^k^t$

On putting the value,

T(0) = 65°F

$65 = Tm + C\\C = 65 - Tm\\\\T = Tm + (65 - Tm) e^k^t$

After 1/2 minute, thermometer reads 110°F. So,

$110 = Tm + (65 - Tm)e^0^.^5^k$                - 1

After 1 minute, thermometer reads 140°F. So,

$140 = Tm + (65 - Tm)e^k$                     - 2

Dividing equation 2 by 1:

$e^k^-^0^.^5^k = \frac{140 - Tm}{110 - Tm} \\\\e^0^.^5^k = \frac{140 - Tm}{110 - Tm}$                                - 3

From 1 we have,

$e^0^.^5^k = \frac{110 - Tm}{65 - Tm}$

Putting this vale in eqn 3. We get,

$\frac{110 - Tm}{65 - Tm} = \frac{140 - Tm}{110 - Tm}$

$(110-Tm)^2 = (140-Tm) (65-Tm)\\\\12100 + Tm^2 - 220Tm = 9100 - 140Tm - 65Tm + Tm^2\\\\3000 - 220Tm = -205Tm\\\\3000 = 15Tm\\\\Tm = 200$

Therefore, the temperature of the oven is 200°F