Question

A pan of warm water (46∘ Celsius) was put in a refrigerator. After 15 minutes, the water's temperature was 27∘ C; 15 minutes after that, it was 19∘C. Use Newton's Law of Cooling to determine how cold the refrigerator was.

Answer 1

The refrigerator was at around 13 ∘C

Explanation:

Newton's Law of Cooling:

The rate of change of a body temperature (amount of heat loss/time of loss) is directly proportional to the difference between its own temperature and the surroundings.

`(T(t2)-T(t1))/(t2-t1) = -h (T(t2) - Tenv)`

T ⇒ temperature

t ⇒ time

Tenv ⇒refrigerator temperature

`(T(t2)-T(t1))/(t2-t1)` ⇒ rate of change of he temperature

-h ⇒ constant of proportionality (negative because the temperature is decreasing inside the refrigerator)

We have 3 points:

time (minutes) - Temperature (∘ C)

0 (when the pan was put in the refrigerator) - 46

15 (after 15 minutes) - 27

30 (15 minutes after the first 15 minutes) - 19

`(27 - 46)/(15 - 0)` = -h (27 - Tenv)

`(19 - 27)/(15 - 0)` = -h (19 - Tenv)

Now we have a system of two equations and two variables

`(-19)/(15) = -h (27 - Tenv)\\ (19)/(15) = h (27 - Tenv)\\ (19)/(15(27 - Tenv)) = h`

`(19 - 27)/(15 - 0) = -h (19 - Tenv)\\(-8)/(15) = -h (19 - Tenv)\\(8)/(15) = h (19 - Tenv)\\(8)/(15) = (19)/(15(27 - Tenv)) (19 - Tenv)\\8(27 - Tenv) = 19 (19 - Tenv)\\216 - 8 Tenv = 361 - 19 Tenv\\19 Tenv - 8 Tenv = 361 - 216\\11 Tenv = 145 \\Tenv = 145/ 11\\Tenv = 13. 18`

The refrigerator was at around 13 ∘C