A roasted turkey is taken from an oven when its temperature has reached 185°F and is placed on a table in a room where the temperature is 51°F.

If the temperature of the turkey is 150°F after half an hour, what is its temperature after 45 min?

When will the turkey cool to 100°F?

Guest Jul 9, 2014

#1**+8 **

As a rough approximation the temperature will fall exponentially. That is the temperature as a function of time is given by:

T(t) = Ta+(T0-Ta)*e^{-k*t}

T(t) is the temperature at time t

T0 is the temperature at time zero (185°F)

Ta is the ambient temperature (51°F)

k is a constant we can find from knowing that the temperature at t=30mins is 150°F

150 = 51 + (185-51)*e^{-k*30}

ln((150-51)/(185-51)) = ln(e^{-k*30})

ln(99/134)=-k*30

k = -ln(99/134)/30 ≈ 0.01 per minute.

So the temperature after 45 minutes is:

T(45) = 51 + 134*e^{-0.01*45} ≈ 136.4°F

Time for turkey to cool to 100°F is given by:

100 = 51 + 134*e^{-0.01*t}

ln(49/134) = -0.01*t

t = -100*ln(49/134) ≈ 100.6 minutes

Alan
Jul 13, 2014

#1**+8 **

Best Answer

As a rough approximation the temperature will fall exponentially. That is the temperature as a function of time is given by:

T(t) = Ta+(T0-Ta)*e^{-k*t}

T(t) is the temperature at time t

T0 is the temperature at time zero (185°F)

Ta is the ambient temperature (51°F)

k is a constant we can find from knowing that the temperature at t=30mins is 150°F

150 = 51 + (185-51)*e^{-k*30}

ln((150-51)/(185-51)) = ln(e^{-k*30})

ln(99/134)=-k*30

k = -ln(99/134)/30 ≈ 0.01 per minute.

So the temperature after 45 minutes is:

T(45) = 51 + 134*e^{-0.01*45} ≈ 136.4°F

Time for turkey to cool to 100°F is given by:

100 = 51 + 134*e^{-0.01*t}

ln(49/134) = -0.01*t

t = -100*ln(49/134) ≈ 100.6 minutes

Alan
Jul 13, 2014