SOLUTION: A hot drink is taken outside on a cold winter day when the air temperature is −7°C. According to a principle of physics called Newton's Law of Cooling, the temperature T (in
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Question 1106913: A hot drink is taken outside on a cold winter day when the air temperature is −7°C. According to a principle of physics called Newton's Law of Cooling, the temperature T (in degrees Celsius) of the drink 't' minutes after being taken outside is given by
T(t) = −7 + Ae^(−kt),
where A and k are constants.
(a)
Suppose that the temperature of the drink is 86°C when it is taken outside. Find the value of the constant A.
(b)
In addition, suppose that 20 minutes later the drink is 29°C. Find the value of the constant k.
(c)
What will the temperature be after 28 minutes?
(d)
When (i.e., after how many minutes) will the temperature reach 0°C?
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
T(t) = −7 + Ae^(−kt)
:
(a) t = 0
86 = -7 + A
A = 93
:
(b) 29 = -7 + 93 * e^(20k)
e^(20k) = 36/93 = 0.3871
20k = ln 0.3871 = −0.9491
k = −0.9491/20 = −0.0475
:
(c) T(t) = -7 + 93 * e^(28*−0.0475) = 17.5964
after 28 minutes 17.5964 C
:
(d) 0 = -7 + 93 * e^(t*−0.0475)
e^(t*−0.0475) = 7/93 = 0.0753
t*−0.0475 = ln 0.0753 = −2.5863
t = −2.5863/−0.0475 = 54.4484 minutes the drink will be 0C
:
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