Algebra word problem help!?
A pizza removed from the oven has a temperature of 427°F. It is left sitting in a room that has a temperature of 68°F. After 18 minutes the temperature of the pizza is 297°F. After how many minutes will the temperature be 122°F? Use T=C+(T (base 0) - C) e^kt
Please help me! I have finals tomorrow & this problem has me stumped!
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I hope this is O.K.
T(n) = C + (T₀ - C) e^(kt)
We are told that the room temperature C = 68.
Thus, when t = 0, e^(kt) = e^0 = 1
427 = 68 + (427 - 68)*1....so, ( T₀ - C) = 359
and when t = 18,
297 = 68 + 359e^(18k)
e^(18k) = 229/359 = 0.637883 = e^-0.4476004
k = -0.4476004/18 = -0.0249778
The temperature fall formula now becomes T(n) = 68 + 359e^(-0.0249778t)
When T(n) = 122 we have :-
122 = 68 + 359e^(-0.0249778t)
e^(-0.0249778t) = 54/359 = 0.15041783 = e^-1.89433832
t = -1.89433832 / -0.0249778 ≈ 76 minutes
If C refers to the room temperature in Centigrade then I'll leave you to do the conversion.