A turkey is taken out of the oven at 17 : 00 and is placed on the kitchen counte

Question

A turkey is taken out of the oven at 17 : 00 and is placed on the kitchen counter. Its initial temperature is 77 C and room temperature is 23 C One hour later, the turkey reaches a temperature of 47 C. Let T (t) be the temperature of the turkey t hours after the initial time. Supposing that the temperature of the turkey follows Newton's Law of Cooling, give the exact formula for T (t) , by finding and solving the corresponding differential equation What is the temperature of the turkey at 19 : 00 ? Give the answer with 0.1 precision. At what time will the temperature of the turkey reach 30 C ? Give the answer as a decimal with 0.1 precision.

Explanation / Answer

(i) Using Newton's Law of Cooling,

dT/dt = k(T – 23).

(ii) Let y(t) = T(t) – 23 , so y(0) = T(0) – 23 = 77– 23= 54,

and so y is a solution of the initial-value problem

               dy/dt = ky with y(0) = 54 and

               dy/y = kdt ,

              log y = kt+c

               y =e k+c

              since y(0) = 54 then

              y =54e kt

            y(60) = 54e60k = 47-23=24

             and so e60k = 24/54=4/9.

              Hence k = (1/60)ln(4/9).

             Thus

               y(t) = 54e(1/60)ln(4/9)t

               y(120) = 54e(120/60)ln(4/9)

               y(120) = 54eln(16/81)

               y(120) = 10.7oC

thus T(120) = 47+10.7=57.7 oC

hence after 2 hours that is at 19:00 the temperature will be 57.7oC.

               

(iii). We solve T(t) = 30

which is the same as y(t) = 30-47 = -17.

From y(t) = 54e(1/60)t ln(4/9) = -17, we get

e(1/60)t ln(4/9) = -17/54

(1/60)t ln (4/9)= ln(-17/54)

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