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

Answer:

The refrigerator was at around 13 ∘C

Step-by-step 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.

$\frac{T(t2)-T(t1)}{t2-t1} = -h (T(t2) - Tenv)$

T ⇒ temperature

t ⇒ time

Tenv ⇒refrigerator temperature

$\frac{T(t2)-T(t1)}{t2-t1}$ ⇒ 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

$\frac{27 - 46}{15 - 0}$ = -h (27 - Tenv)

$\frac{19 - 27}{15 - 0}$ = -h (19 - Tenv)

Now we have a system of two equations and two variables

$\frac{-19}{15} = -h (27 - Tenv)\\ \frac{19}{15} = h (27 - Tenv)\\ \frac{19}{15(27 - Tenv)} = h$

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

The refrigerator was at around 13 ∘C