Question

Question 4 3 pts Newton's Law of Cooling states that the rate of change of the temperature of a cooling body is proportional to the difference between the temperature of the body and the constant temperature of the medium surrounding the body, Apply this law to the following problem: At 10:00 am, a woman took a cup of hot instant coffee from her microwave oven and placed it on a nearby kitchen counter to cool. At this instant the temperature of the coffee was 180°F, and 10 minutes later it was 160°F. If we assume that the constant temperature of the kitchen was 70°F, what was the temperature of the coffee at 10:15 am? 27/11 10+ F 7(10 (19) 20(vT+) 7(10 115) F o 7/11 10+ F O None of them 2711 101 7+ 11

Answer 1

Solution:

Let T is the coffee temperature at time t , and Tk is the kitchen temperature.

Then , assuming Newton's Law of Cooling, we have--->

dT/dt = k (T - Tk)

Then, dT/dt = k(T-70)

Separating Variables, we obtain,

dT/(T-70) = kdt

By integrating and simplifying , we then see that

ln(T-70) = kt + c1

T-70 = exp(kt) + exp(c1)

Therefore, T = 70+c* exp(kt)

To get constant c , apply the initial condition , at t = 0 yrs and T =180 degree F , Then we obtain,

180 = 70 +c*exp(0)

Or, c = 110

Then, we have, T = 70+110*exp(kt).

To get constant k , apply the condition t =10 mins and T = 160 degree F , Then we obtain,

160= 70 + 110 exp(10 k )

Or, 90 = 110 exp(10k)

k = -0.02

Then, we have, T = 70 + 110 exp(-0.02t)

Now putting , t =15 mins , we obtain

T = 70 + 110 exp(-0.02*15) = 151.4 degree F = 10[7+27*root(11) /11 ] degree F ( optionD)