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) About Me  (Show Source):
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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 11.2%2F3°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) About Me  (Show Source):
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Newton's law of cooling states that
dT%2Fdt+=+-k%28T-T%5Ba%5D%29,

where T is the temperature of the object and T%5Ba%5D is the temperature of the surrounding.
==>dT%2F%28T-75%29+=+-k%2Adt, or ln%28T-75%29+=+-kt+%2B+c after integrating both sides wrt t.
Now T(0) = 185 ==> ln(185 - 75) = -k*0 + c <==> c = ln(110) ==> ln%28%28T-75%29%2F110%29+=+-kt

T(30) = 150 ==> ln%28%28150-75%29%2F110%29+=+-30k, or %28-1%2F30%29%2Aln%2815%2F22%29+=+k
==> k++=+%281%2F30%29%2Aln%2822%2F15%29
==> ln%28%28T-75%29%2F110%29+=+-%281%2F30%29%2Aln%2822%2F15%29%2At , or
==> %28T-75%29%2F110+=+e%5E%28-%281%2F30%29%2Aln%2822%2F15%29%2At%29
After t = 50 minutes, %28T-75%29%2F110+=+e%5E%28-%2850%2F30%29%2Aln%2822%2F15%29%29
==> highlight%28T+=+133.1%5E0%29F
The roast turkey will cool to 105 degrees F after
ln%28%28105-75%29%2F110%29+=+-%281%2F30%29%2Aln%2822%2F15%29%2At, or highlight%28t+=+-%2830%2Aln%283%2F11%29%29%2Fln%2822%2F15%29+=+101.77%29 minutes.

Answer by ikleyn(52756) About Me  (Show Source):
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.
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) = 75+%2B+%28185-75%29%2Ae%5E%28-kt%29 = 75+%2B+110%2Ae%5E%28-kt%29.


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) = 75%2F110 = 0.6818

    - 30k = ln(0.6818)

    k = -ln%28%280.6818%29%29%2F30 = 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 = 75+%2B+110%2Ae%28-0.01277t%29


    e(-0.01277*t) = %28105-75%29%2F110 = 0.2727

    -0.01277*t = ln(0.2727)

    t = -ln%280.2727%29%2F0.01277 = 102 minutes   (rounded).       ANSWER to question (b)

Solved.



Answer by MathTherapy(10549) About Me  (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: t is the time at a COOLED temperature 
                                                                        T%28t%29 is the TEMPERATURE (T) at a given time (t) 
                                                                        T%5Bs%5D is the SURROUNDING temperature 
                                                                        T%5Bo%5D is the ORIGINAL/INITIAL temperature 
                                                                        +k is the CONSTANT or COOLING rate

 then becomes: matrix%281%2C3%2C+T%2830%29%2C+%22=%22%2C+75+%2B+%28185+-+75%29e%5E%28-+30k%29%29 -- Substituting 30 for t, 75 for T%5Bs%5D, and 185 for T%5Bo%5D
                                           matrix%281%2C3%2C+150%2C+%22=%22%2C+75+%2B+%28185+-+75%29e%5E%28-+30k%29%29 -- Substituting 150 for T%2830%29
                                            
                                           matrix%281%2C3%2C+-+30k%2C+%22=%22%2C+ln%2815%2F22%29%29 ----- Converting to LOGARITHMIC (Natural) form 

                    CONSTANT/COOLING RATE, or 

 becomes:  -- Substituting 75, 185, - .012766408, and 50  
                                                                            for T%5Bs%5D, T%5Bo%5D, k, and t, respectively
                                     matrix%281%2C3%2C+T%2850%29%2C+%22=%22%2C+75+%2B+110e%5E%28-+.012766408%2850%29%29%29
   Temperature after 50 minutes, or 

 becomes: matrix%281%2C3%2C+105%2C+%22=%22%2C+75+%2B+%28185+-+75%29e%5E%28-+.012766408t%29%29%29 -- Substituting 105, 75, 185, and - .012766408,                                   
                                                                           for T%28t%29, T%5Bs%5D, T%5Bo%5D, and k, respectively
                                      matrix%281%2C3%2C+105%2C+%22=%22%2C+75+%2B+110e%5E%28-+.012766408t%29%29 
                                       
                                      matrix%281%2C3%2C+-+.012766408t%2C+%22=%22%2C+ln%283%2F11%29%29 ----- Converting to LOGARITHMIC (Natural) form 

TIME taken for turkey to cool to 105o, or