# A thermometer is taken from a room where the temperature is 23 ^o C to the outdoors, where the temperature is -15 ^o C. After one minute the thermometer reads 11 ^o C. ?

## A thermometer is taken from a room where the temperature is 23 ^o C to the outdoors, where the temperature is -15 ^o C. After one minute the thermometer reads 11 ^o C. (a) What will the reading on the thermometer be after 5 more minutes? , (b) When will the thermometer read -14 ^o C? minutes after it was taken to the outdoors.

Jan 26, 2018

(a) - $- {11.1}^{\circ} C$
(b) - after $575$ seconds i.e. $8$ minutes and $35$ seconds temperature would be $- {14}^{\circ} C$

#### Explanation:

According to Newton's Law of Cooling, the rate of the cooling is proportional to the temperature difference between the object and its surroundings.

In other words, if an object at a higher temperature say ${T}_{0}$ is moved to a surrounding at a lower temperature ${T}_{s}$, the rate of cooling is directly proportional to the difference in temperature of te two i.e. $\frac{\mathrm{dT}}{\mathrm{dt}} = - k \left({T}_{0} - {T}_{s}\right)$. This means that temperature ${T}_{t}$ of the object at $t$ time is given by ${T}_{t} = {T}_{s} + \left({T}_{0} - {T}_{s}\right) {e}^{- k t}$

Here we have ${T}_{0} = {23}^{\circ}$, ${T}_{s} = - {15}^{\circ}$ and at $t = 60$, we have ${T}_{60} = {11}^{\circ}$ - here we have used $60$ as we intend to use time in seconds.

Hence $11 = - 15 + \left(23 - \left(- 15\right)\right) {e}^{- 60 k}$ or $26 = 38 {e}^{- 60 k}$

or ${e}^{- 60 k} = \frac{26}{38}$ and hence

$k = - \ln \frac{\frac{26}{38}}{60} = 0.00632482702$

(a) After $5$ more minutes i.e. after $360$ seconds, temperature would be

${T}_{360} = - 15 + 38 {e}^{- 360 \times 0.00632482702}$

= $- 15 + 3.9 = - {11.1}^{\circ}$

(b) Let the temperature be $- {14}^{\circ} C$ after $t$ seconds, then

$- 14 = - 15 + 38 {e}^{- 0.00632482702 t}$

or ${e}^{- 0.00632482702 t} = \frac{15 - 14}{38} = 0.026315789$

i.e. $t = - \ln \frac{0.026315789}{0.00632482702} \cong 575$

Hence after $575$ seconds i.e. $8$ minutes and $35$ seconds temperature would be $- {14}^{\circ} C$