Question

Newton’s law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton’s law of cooling. If the coffee has a temperature of 200◦F when freshly poured, and 1 min later has cooled to 190◦F in a room at 70◦F, determine when the coffee reaches a temperature of 150◦F.

Answer 1

6 mins

Explanation:

Given that;

T(t) = Ce^-kt + T(s)

Since T(t) = 200◦F, and T(s) = 70◦F where t = 0

200= Ce^-k(0) + 70

C= 200 - 70

C = 130

Then when T(t) = 190◦F and t= 1s ;

190 = 130e^-(k *1) + 70

190 - 70 = 130e-^(k *1)

120/130 = e^k

0.923 = e^-k

-k = ln(0.923)

k = 0.08

To determine the time taken to reach a temperature of 150◦F

150 = 130e^-(0.08t) + 70

150 - 70/130 = e^-(0.08t)

0.6154 = e^-(0.08t)

-(0.08t) = ln 0.6154

0.08t = 0.4855

t = 0.4855/0.08

t = 6 mins