SOLUTION: Suppose that a cup of hot coffee at a temperature of T0 is set down to cool in a room where the temperature is kept at T1. Then the temperature of the coffee as it cools can be mod

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Question 1043870: Suppose that a cup of hot coffee at a temperature of T0 is set down to cool in a room where the temperature is kept at T1. Then the temperature of the coffee as it cools can be modeled by the function . Here, f(t) = (T0 - T1)ekt + T1 is the temperature of the coffee after t minutes; t=0 corresponds to the initial instant when the temperature of the coffee is T0; and k is a (negative) constant that depends, among other factors, on the dimensions of the cup and the material from which it is constructed. [We get this formula in calculus and is based on Newton's law of cooling.] Here are the two problems I want to discuss this week.
Problem A: Suppose that a cup of hot coffee at a temperature of 185°F is set down to cool in a room where the temperature is kept at 70F°. What is the temperature of the coffee ten minutes later? Use k = -0.15
Problem B: Suppose that a cup of hot coffee at a temperature of 185°F is set down to cool in a room where the temperature is kept at 70°F. How long will it take for the coffee to cool to 140°F?
Answer by advanced_Learner(501) About Me  (Show Source):
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Suppose that a cup of hot coffee at a temperature of T0 is set down to cool in a room where the temperature is kept at T1. Then the temperature of the coffee as it cools can be modeled by the function . Here, f(t) = (T0 - T1)ekt + T1 is the temperature of the coffee after t minutes; t=0 corresponds to the initial instant when the temperature of the coffee is T0; and k is a (negative) constant that depends, among other factors, on the dimensions of the cup and the material from which it is constructed. [We get this formula in calculus and is based on Newton's law of cooling.] Here are the two problems I want to discuss this week.
Problem A: Suppose that a cup of hot coffee at a temperature of 185°F is set down to cool in a room where the temperature is kept at 70F°. What is the temperature of the coffee ten minutes later? Use k = -0.15
Problem B: Suppose that a cup of hot coffee at a temperature of 185°F is set down to cool in a room where the temperature is kept at 70°F. How long will it take for the coffee to cool to 140°F

a.
T0=185,T1=70,K=-0.15,t=time=10 min,
Modelling equation is
f%28t%29+=+%28T0+-+T1%29e%5E%28k%2At%29+%2B+T1
A
find t=10
f%28t=10%29 = %28185+-+70%29%2Ae%5E%28-0.15%2A10%29+%2B+70
f%28t=10%29+= %28115%29%2Ae%5E%28-1.5%29+%2B+70
f%28t=10%29+= %28115%29%2Ae%5E%28-1.5%29+%2B+70
f%28t=10%29+= %28115%29%2A%280.223%29+%2B+70
f%28t=10%29+= 25.625+%2B+70
f%28t=10%29 = 95.625
This means that when time is 10 min the temperature is 95.625 which has cooled from 185.

B
140+= %28185+-+70%29%2Ae%5E%28-0.15%2At%29+%2B+70
140+= %28115%29%2Ae%5E%28-0.15%2At%29+%2B+70
70+= %28115%29%2Ae%5E%28-0.15%2At%29+
70+= %28115%29%2Ae%5E%28-0.15%2At%29+
%2870%29%2F%28115%29 = e%5E%28-0.15%2At%29
%2814%2F23%29 = e%5E%28-0.15%2At%29
Take Ln of both sides
%28Ln%2814%2F23%29%29 =lne%5E%28-0.15%2At%29%29
%28Ln%2814%2F23%29%29 =%28-0.15t%2Alne%29
ln%28e%29=1
%28Ln%2814%2F23%29%29 =-0.15t%2A%281%29
%28Ln%2814%2F23%29%29%2F%28-0.15%29+=t+
%28Ln%2814%2F23%29%29%2F%28-0.15%29 =t+
%28-0.496%29%2F%28-0.15%29=t
t=3.3+

time is 3.3
it will 3.3 mins to cool to 140.