A pan of water (46 degrees) was put into a fridge. Ten min. later, the temp was 39; 10 min. later, the temp was 33.

Use newtons law of cooling to find the temp of the fridge..

I know the equation.

Do I need to find the rate, or how do I do this?

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- Dec 3rd 2009, 07:00 PMMorgan82find temp
A pan of water (46 degrees) was put into a fridge. Ten min. later, the temp was 39; 10 min. later, the temp was 33.

Use newtons law of cooling to find the temp of the fridge..

I know the equation.

Do I need to find the rate, or how do I do this? - Dec 3rd 2009, 08:16 PMnehme007
By Newton's law of cooling you have that the temperature of the water is given by:

$\displaystyle T(t) = T_f + (46 - T_f) e^{-rt} $

where $\displaystyle T_f $ is the temperature of the fridge and r is some unknown rate constant. You are told T(10) = 39 and T(20) = 33, so:

$\displaystyle T(10) = 39 = T_f + (46 - T_f) e^{-10r} $

$\displaystyle T(20) = 33 = T_f + (46 - T_f) e^{-20r} $

The first equation gives you:

$\displaystyle r = -1/10 ln(\frac{39-T_f}{46 - T_f}) $

and the second gives you:

$\displaystyle r = -1/20 ln (\frac{33-T_f}{46 - T_f}) $.

Equating these two expressions for r and multiplying by -20 yields:

$\displaystyle 2ln(\frac{39-T_f}{46 - T_f}) = ln (\frac{33-T_f}{46 - T_f}) $.

The 2 can move inside the log as a power, then you can get rid of the logs and solve. You should get $\displaystyle T_f = -3 $.