The rate at which a body cools also depends on its exposed surface area S. If S is a constant, then a modification of (2), given
in Section 3.1, is dT dt = kS(T − Tm), where k < 0 and Tm is a constant. Suppose that two cups A and B are filled with coffee at the same time. Initially, the temperature of the coffee is 145° F. The exposed surface area of the coffee in cup B is twice the surface area of the coffee in cup A. After 30 min the temperature of the coffee in cup A is 95° F. If Tm = 65° F, then what is the temperature of the coffee in cup B after 30 min? (Round your answer to two decimal places.)
1 answer:
Answer:
76.25°
Step-by-step explanation:
The solution to the differential equation is an exponential curve with a horizontal asymptote at Tm. It passes through (0, 145) and (30, 95), so the equation can be written as ...
T = 80 +65((95-65)/(145-65))^(t/30)
T = 80 +65(3/8)^(t/30)
That is, the temperature difference is reduced to 3/8 of its original value in 30 minutes.
Since the coffee in cup B cools twice as fast, it will cool to the same temperature (95°) in 15 minutes. In the next 15 minutes, the temperature difference will be reduced to (3/8)^2 of the original 80°, so will be 11.25°. That is, the temperature of cup B will be ...
11.25° +65° = 76.25°
after 30 minutes.
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Hope you could get an idea from here.
Doubt clarification - use comment section.
