SOLUTION: A roast 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 75°F. (Round your answers to the nearest w

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Question 1178344: A roast 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 75°F. (Round your answers to the nearest whole number.)
(a) If the temperature of the turkey is 150°F after half an hour, what is the temperature after 50 minutes?
T(50) =

(b) When will the turkey have cooled to 105°?
Found 4 solutions by Lol_dude, robertb, ikleyn, MathTherapy:
Answer by Lol_dude(1)   (Show Source): You can put this solution on YOUR website!
This question relies on the linear decrease in temperature. When it was first taken out of the oven, or at T(0), it is 185°F. After 30 mins, it is now 150°F. Therefore, after 30 mins, it has decreased by 35°F. Every ten minutes, it decreases by 11.°F. At T(50), which is 20 minutes later, it would of cooled by 23.3333...°F.
a) The temperature at T(50) would be 126.666...°F.
The difference between 185°F and 105°F is 80°F. Every one minute, the turkey cools by 1.166...°F.
Answer by robertb(5830)   (Show Source): You can put this solution on YOUR website!
Newton's law of cooling states that
,

where T is the temperature of the object and is the temperature of the surrounding.
==>, or after integrating both sides wrt t.
Now T(0) = 185 ==> ln(185 - 75) = -k*0 + c <==> c = ln(110) ==>

T(30) = 150 ==> , or
==>
==> , or
==>
After t = 50 minutes,
==> F
The roast turkey will cool to 105 degrees F after
, or minutes.
Answer by ikleyn(52765)   (Show Source): You can put this solution on YOUR website!
.
A roast 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 75°F. (Round your answers to the nearest whole number.)
(a) If the temperature of the turkey is 150°F after half an hour, what is the temperature after 50 minutes?
T(50) =
(b) When will the turkey have cooled to 105°?
~~~~~~~~~~~~~~~~~~ 


The Newton law of cooling states that the temperature of the roast turkey in the room is this function of time t


    T(t) =  = .


where "k" is the decay constant.  

At t= 30 minutes  T(t)= 150°F,  which gives you an equation to find the decay constant k:


    150 = 75 + 110*e^(-k*30)

    110*e^(-k*30) = 150 - 75 = 75

    e^(-30k) =  = 0.6818

    - 30k = ln(0.6818)

    k =  = 0.01277.


Thus the decay constant k is found.


        Now I am in position to answer questions (a) and (b).


The temperature after 50 minutes is


    T(50) = 75 + 110*e^(-0.01277*50) = 75 + 110*2.71828^(-0.6385) = 133°F.      ANSWER to question (a)



To find the time getting 105°F, use and solve this equation 


    105 = 


    e(-0.01277*t) =  = 0.2727

    -0.01277*t = ln(0.2727)

    t =  = 102 minutes   (rounded).       ANSWER to question (b)

Solved.



Answer by MathTherapy(10549)   (Show Source): You can put this solution on YOUR website!
A roast 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 75°F. (Round your answers to the nearest whole number.)
(a) If the temperature of the turkey is 150°F after half an hour, what is the temperature after 50 minutes?
T(50) =

(b) When will the turkey have cooled to 105°?
Formula for Newton's Law of cooling:  where:  is the time at a COOLED temperature 
                                                                         is the TEMPERATURE (T) at a given time (t) 
                                                                         is the SURROUNDING temperature 
                                                                         is the ORIGINAL/INITIAL temperature 
                                                                         is the CONSTANT or COOLING rate

 then becomes:  -- Substituting 30 for t, 75 for , and 185 for 
                                            -- Substituting 150 for 
                                            
                                            ----- Converting to LOGARITHMIC (Natural) form 

                    CONSTANT/COOLING RATE, or 

 becomes:  -- Substituting 75, 185, - .012766408, and 50  
                                                                            for , , k, and t, respectively
                                     
   Temperature after 50 minutes, or 

 becomes:  -- Substituting 105, 75, 185, and - .012766408,                                   
                                                                           for , , , and k, respectively
                                       
                                       
                                       ----- Converting to LOGARITHMIC (Natural) form 

TIME taken for turkey to cool to 105o, or

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