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?
+33615 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
+33615 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
