SOLUTION: cup of coffee at 85°C is placed in a freezer at 0°C.The temperature of the coffee decreases exponentially, so that after 5 minutes it is 30°C. (a)What is its temperature after

Question 1163608: cup of coffee at 85°C is placed in a freezer at 0°C.The temperature of the coffee decreases exponentially, so that after 5 minutes it is 30°C.
(a)What is its temperature after 3 minutes?
(b)Find by trial how long it will take for the temperature to drop ton5°C
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
one of the formula that works best with these types of problem is:
f = p * (1 + r) ^ n
f is the future value
p is the present value
r is the interest rate per time period
n is the number of time periods.

in your problem, the equation becomes:
30 = 85 * (1 + r) ^ 5
divide both sides of the problem by 85 and then take the fifth root of both sides of the equation to get:
(30/85)^(1/5) = 1 + r
subtract 1 from both sides of the equation to get:
(30/85)^(1/5) - 1 = r
solve for r to get:
r = -.1880290984
replace r with that in the original equation to get:
30 = 85 * (1-.1880290984) ^ 5 = 30, confirming the value of r is good.
now that you know r, you can solve the equation.
the equation becomes:
5 = 85 * (1-.1880290984) ^ n
divide both sides of the equation by 85 and take the log of both sides of the equation to get:
log(5/85) = log(1-.1880290984)^n)
since loc(x^a) = a * log(x), this equation becomes:
log(5/85) = n * log(1-.1880290984)
divide both sides by log(1-.1880290984) to get:
log(5/85)/log(1-.1880290984) = n
solve for n to get:
n = 13.60220271 minutes
it will take 13.60220271 minutes for the temperature to drop from 85 degrees to 5 degrees.
that was the answer to your question b.
the answer to your wuestion a follows.
after 3 minutes, the equation becomes:
f = 85 * (1-.1880290984)^3 = 45.50303066 degrees.
the equation can be graphed.
it is shown below:

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