Newton's law says that the rate of cooling is proportional to the difference between the ambient temperature and the temperature of the object. Let the room temp T_R and the temperature of the kettle at time t, T(t)

dT/dt = k(T - T_R), T(0) = 100, T(15) = 75, T_R = 20,

T(t) = Ae^(kt) + 20, where A is a constant, k is the heat transfer coefficient

a. Find the heat transfer coefficient k, and find a model for the temperature T of the water t hours after it is brought to a boil.

T(t) = Ae^(kt) + 20

for t = 0 we have T(0) = 100

A + 20 = 80 => A = 60

for t = 15 we have T(15) = 75

A*e^(15k) + 20 = 75

60*e^(15k) = 75 - 20

e^(15k) = 55/60

k = (ln(55/60))/15

k = - 0.0058 in the heat transfer coefficient

model for the temperature T of the water t hours after it is brought to a boil

T(t) = 60*e^(- 0.0058t) + 20

b.Use the model to predict the temperature of the water after 25 minutes

for t = 25 we have

T(25) = 60*e^(- 0.0058*25) + 20

T(25) = 70.901 degrees

the temperature of the water after 25 minutes will be 70.901 degrees

c.How long will it take the water to cool to 40◦C?

t = ? if T(t) = 40

60*e^(- 0.0058*t) + 20 = 40

e^(- 0.0058*t) = 20/60

t = (- ln(20/60))/0.0058

t = 189.416 minutes = 3 hours 9.416 min

it takes 3 hours 9.416 min for the water to cool to 40 degrees.