Question

According to Newton's Law of Cooling, the temperature T(t) of a hot object, at time t after being placed in an environment with a constant temperature C, is given by T(t)=C+(To-C)e^-kt, where To is the temperature of the object at time t = 0 and k is a constant that depends on the object. If a hot cup of coffee, initially at 190°F, cools to 125°F in 5 minutes when placed in a room with a constant temperature 75°F, how long will it take for the coffee to reach 100°F?​

Answer 1

Given the temperature of an object at time t as

-kt f(t) = T. + Ce +

where TO is the environment temperature, C is the difference in temperature of environment and object.

Given a boliling water at 100 degree celsius is placed in a freezer at 0 degree celsius, so we have

To = 0

C = 100 – 0= 100

After 26 minutes the temperature of water is 50 degree celsius.

f(26) = 0 + 100e -26k = 50

100e 26K = 50

-26 e = 0.5

-26%: In e = ln 0.5

-26% = ln 0.5

In 0.5 k= 26

So the temperature of water after 78 minutes will be,

f(78) = 100e -78 In 0.5 26 = = 12.5 =

The temperature of water after 78 minutes will be approximately 12.5 0 C

Explanation: