Question

Knowing `T-T_s=(T_0 - T_s)e^(kt)`, A pan of warm water (46dgC) was put in a refrigerator. 10 minutes later, the water's temperature was 39dgC; 10 minutes after that, it was 33dgC. Use Newton's law of cooling to estimate how cold the refrigerator was?

The question is asking for `T_s`.

Answer 1

I got `-3^@ "C"`.

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Well, first, let's define what we have.

• `T` is the current water temperature.
• `T_0 = 46^@ "C"` is the starting water temperature.
• `T_s` is the temperature of the fridge (the surroundings).
• `t` is the time passed in minutes.
• `k` is the rate constant of the cooling process, constant with respect to the surrounding temperature.

Also, there's a typo... should be `e^(-kt)`, since `T - T_s` will decrease over time, but `e^(kt)` will increase over time (`T_0 - T_s` is constant).

So, what we really have is:

`T - T_s = (46 - T_s)e^(-kt)`

From the two data points, which are `(T_1,t_1) = (39, 10)` and `(T_2,t_2) = (33, 20)`, we form two equations:

`39 - T_s = (46 - T_s)e^(-10k)`
`33 - T_s = (46 - T_s)e^(-20k)`

Dividing these equations gives:

`(39 - T_s)/(33 - T_s) = e^(-10k + 20k) = e^(10k)`

From this, the rate constant in terms of `T_s` is:

`k = 1/10ln((39 - T_s)/(33 - T_s))`

Plugging this back into the original equation:

`T - T_s = (46 - T_s)e^(-t/10ln((39 - T_s)/(33 - T_s)))`

Now, suppose `10` minutes passed. If `T_s > 39`, it doesn't make sense.

`39 - T_s = (46 - T_s)e^(-10/10ln((39 - T_s)/(33 - T_s)))`

`(39 - T_s)/(46 - T_s) = e^(-ln((39 - T_s)/(33 - T_s)))`

`= e^(ln((33 - T_s)/(39 - T_s)))`

`= (33 - T_s)/(39 - T_s)`

Solving this now, we get:

`(39 - T_s)^2 = (33 - T_s)(46 - T_s)`

`1521 - 78T_s + T_s^2 = 1518 - 79T_s + T_s^2`

`1521 + T_s = 1518`

`color(blue)(T_s = -3^@ "C")`