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

###### Question:

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