Daniel decides he wants to make a stew, too, but decides to cook his in tge oven at a higher temperature of 350 F. They are in the same room, so the temperature is the same. If it takes Daniel's stew 50 minutes to cool to serving temperature,what is his cooling rate??? PLEASE HELP (Newton's law of cooling)!!!!!

Question

whpalmer4 · Answer

Is this part of another problem?  We don't seem to have all of the information here.

anonymous · Answer

Its the entire problem

whpalmer4 · Answer

What is the serving temperature?  "They are in the same room" — what is "they"?

anonymous · Answer

That's what the problem says ,, i have no idea. Isn't room temperature 70F?

whpalmer4 · Answer

But you don't typically eat your food at room temperature.

whpalmer4 · Answer

Well, I don't :-)

whpalmer4 · Answer

Well, Newton's law of cooling says that the rate of cooling is proportional to the difference in temperature between the object which is cooling and the ambient temperature.

Let's call the temperature of the stew at time $t$ simply $T(t)$
Let's call the ambient temperature (whatever it is) $T_a$
The difference in temperature will be $T(t) - T_a$
We'll call that $y(t)$ to make things a bit neater.  $y(t) = T(t)-T_a$

okay so far?

whpalmer4 · Answer

We'll have an initial temperature of $T(0) = T_0 = 350^\circ\text{F} = 449.8^\circ\text{K}$