Newton’s law of cooling states that the rate of cooling of an object is proportional to the temperature

5 answers
Question:

Newton’s law of cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. so the rate of cooling for a bottle of lemonade at a room temperature of 75°f which is placed into a refrigerator with temperature of 38°f can be modeled by dt/dt=k(t-38) where t(t) is the temperature of the lemonade after t minutes and t(0) = 75. after 30 minutes the lemonade has cooled to 60°f, so t(30) = 60.
to the nearest degree, what is the temperature of the lemonade after an additional 30 minutes?

Answers

DT/dt = k(T - 38)
dT/(T - 38) = kdt
∫dt/(T - 38) = k∫dt
ln(T - 38) = kt + c
T - 38 = e^(kt + c) = Ae^kt; where A = e^c
T = Ae^kt + 38
T(0) = Ae^0 = A = 75
T(30) = 75e^30k = 60
e^30k = 60/75 = 0.8
30k = ln(0.8)
k = ln(0.8) / 30

T(60) = 75e^(60(ln(0.8) / 30)) = 75e^(2ln(0.8)) = 75(0.64) = 48 degrees.

NO!!! answer is 51 !

Step-by-step explanation:

https://answers.yahoo.com/question/index?qid=20180507053559AAFm7NV

a) The initial temperature difference of 75-38 = 37 degrees has been reduced to 60-38 = 22 degrees in 30 minutes. According to the law of cooling, it will be reduced by the same factor of 22/37 in another 30 minutes, so the temperature difference then will be 22(22/37) = 13 degrees.

The temperature of the lemonade after an additional 30 minutes will be 38+13= 51°F.

b) The temperature can be modeled by
  T(t) = 38 + 37·(22/37)^(t/30)
Substituting the given information, we have
  T(t) = 55 = 38 +37·(22/37)^(t/30)
  17/37 = (22/37)^(t/30) . . . . . . subtract 38, divide by 37
  log(17/37) = (t/30)·log(22/37) . . . . take the logarithm
  t = 30·log(17/37)/log(22/37) . . . . divide by the coefficient of t
  t ≈ 44.878

It will take about 45 minutes for the lemonade to cool to 55 °F.
[tex]Newton’s law of cooling states that the rate of cooling of an object is proportional to the temperat[/tex]

After the first 30 minutes,
dT = 75-60
= 15
dt = 30 mins
k = (15/60) * (60 - 38)
k = 5.5
After 60 minutes,
dT = 60 - T
dt = 30
(60 - T)/30 = 5.5(T - 38)
60 - T = 165T - 6270
166T = 6330
T = 38.13°F

DT/dt=k(T-38)
ln(T-38) = kt+C
T=38 + Ce^(-kt)
Since T(0)=75, 75 = 38+C, so C=37
Since T(30)=60, 60 = 38+37e^(-30k), so k = -ln(22/37)/30 ~= -0.017329
T(60) = 38+37e^(-0.017329*60) = 51.1°F

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