Question

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.

Answer 1

(a) - `-11.1^@C`
(b) - after `575` seconds i.e. `8` minutes and `35` seconds temperature would be `-14^@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. `(dT)/(dt)=-k(T_0-T_s)`. This means that temperature `T_t` of the object at `t` time is given by `T_t=T_s+(T_0-T_s)e^(-kt)`

Here we have `T_0=23^@`, `T_s=-15^@` and at `t=60`, we have `T_60=11^@` - here we have used `60` as we intend to use time in seconds.

Hence `11=-15+(23-(-15))e^(-60k)` or `26=38e^(-60k)`

or `e^(-60k)=26/38` and hence

`k=-ln(26/38)/60=0.00632482702`

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

`T_360=-15+38e^(-360xx0.00632482702)`

= `-15+3.9=-11.1^@`

(b) Let the temperature be `-14^@C` after `t` seconds, then

`-14=-15+38e^(-0.00632482702t)`

or `e^(-0.00632482702t)=(15-14)/38=0.026315789`

i.e. `t=-ln0.026315789/0.00632482702~=575`

Hence after `575` seconds i.e. `8` minutes and `35` seconds temperature would be `-14^@C`