Newton's Law of Cooling

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q_fruit

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In Newton's Law of Cooling,
T = T<sub>s</sub> + (T<sub>0</sub> - T<sub>s</sub>)e<sup>-kt</sup>

but I can't figure out how to find the surrounding temperature (T<sub>s</sub>) when you're only told that "A pan of 46<sup>o</sup>C water was put in a fridge. 10 minutes later, the water's temp was 39<sup>o</sup>C; 10 minutes after that, it was 33<sup>o</sup>C. Use Newton's law of cooling to estimate how cold the fridge was."

So I know that T<sub>o</sub>=46 and t=10 minutes...I'm trying to find K first, but I don't know if that's the correct first step.

Thanks in advance.
 
It might help if you said what the variables stand for...?

Thank you.

Eliz.
 
sorry.

T = temp at time t
T<sub>0</sub> = initial temp
T<sub>s</sub> = surrounding temp
 
Thank you for clarifying. Some formulas are sufficiently standardized and common as to be recognizable, but I was having difficulty with this one.

Okay, so you've got the initial situation:

. . . . .T<sub>0</sub> = 46 [fixed]
. . . . .T = 39 at t = 10
. . . . .T = 33 at t = 20

Then, plugging into the equation:

. . . . .39 = T<sub>s</sub> + (46 - T<sub>s</sub>)e<sup>-10k</sup>

. . . . .33 = T<sub>s</sub> + (46 - T<sub>s</sub>)e<sup>-20k</sup>

This gives you two equations in two unknowns. Solve for T<sub>s</sub> and k.

Actually, you can probably do this without ever finding the value of k: Let e<sup>-10k</sup> = D; then e<sup>-20k</sup> = e<sup>2(-10k)</sup> = [e<sup>-10k</sup>]<sup>2</sup> = D<sup>2</sup>. Solve each equation for the e-part, and use the fact that the expression for e<sup>-20k</sup> is equal to the square of the expression for e<sup>-10k</sup> to solve for T<sub>s</sub>.

It's a little messy. If you get stuck, please reply showing your steps. Thank you.

Eliz.

Note: I don't get this being a "fridge"; it computes as a freezer for me.
 
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