(1 point) Newton\'s Law of Cooling states that the rate at which an object cools

Question

(1 point) Newton's Law of Cooling states that the rate at which an object cools is proportional to the difference in temperature between the object and the surrounding medium. Thus, if an object is taken from an oven at 307 F and left to cool in a room at 76°F, its temperature T after t hours will satisfy the differential equation dT = k(T-76). dt If the temperature fell to 190° F in 0.5 hour(s), what will it be after 4 hour(s)? After 4 hour(s), the temperature will be Hint: Newton's Law of Cooling is discussed in the book on pages 240-241. degrees F.

Explanation / Answer

Solution:

dT / dt = k(T - 76)
dT / (T - 76) = k dt ; Integrate both sisdes
ln (T - 76) = kt + c

convert to exponential:

e(kt + c) = T - 76
Cekt = T - 76
T = 76 + Cekt

we know that T(0) = 307 ; So

307 = 76 + C(1) ; {Let ekt = e0 = 1}
C = 307 - 76
C = 231

T = 76 + 231 ekt

Another data point: T(0.5) = 190
subtitute to find k:

190 = 76 + 231 e0.5k
114 = 231 e0.5k
114 / 231 = e0.5k
ln (114 / 231) = 0.5k
ln (114 / 231) / 0.5 = k

Now we know that T(t) = 76 + 231 et{ln(114/231) / 0.5}

Plug in t = 4 to find T(4), or the temperature after 4 hours;

T(4) = 76 + 231 e4{ln(114/231) / 0.5}

T(4) = 76 + 231 e8{ln(114/231)}

T(4) = 76 + 231 e{ln(114/231)^8}

T(4) = 76 + 231 * (114/231)^8

T(4) = 76.81 degree F

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