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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

Best Answer 

 #1
avatar+26399 
+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
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1+0 Answers

 #1
avatar+26399 
+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

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