The answer is 35 minutes
The Newton's law of cooling is:
T(x) = Ta + (To - Ta)e⁻ⁿˣ
T(x) - the temperature of the coffee at time x
Ta - the ambient temperature
To - the initial temperature
n - constant
step 1. Calculate constant k:
We have:
T(x) = 200°F
x = 10 min
Ta = 68°F
To = 210°F
n = ?
T(x) = Ta + (To - Ta)e⁻ⁿˣ
200 = 68 + (210 - 68)e⁻ⁿ¹⁰
200 = 68 + 142 e⁻¹⁰ⁿ
200 - 68 = 142 e⁻¹⁰ⁿ
132 = 142 e⁻¹⁰ⁿ
e⁻¹⁰ⁿ = 132/142
e⁻¹⁰ⁿ = 0.93
Logarithm both sides with natural logarithm:
ln(e⁻¹⁰ⁿ) = ln(0.93)
-10n ln(e) = -0.07
-10n 1 = - 0.07
-10n = -0.07
n = -0.07 / - 10
n = 0.007
Step 2. Calculate time x when T(x) = 180°F:
We have:
T(x) = 180°F
x = ?
Ta = 68°F
To = 210°F
n = 0.007
T(x) = Ta + (To - Ta)e⁻ⁿˣ
180 = 68 + (210 - 68)e⁻⁰.⁰⁰⁷ˣ
180 - 68 = 142 e⁻⁰.⁰⁰⁷ˣ
112 = 142 e⁻⁰.⁰⁰⁷⁾ˣ
e⁻⁰.⁰⁰⁷ˣ = 112/142
e⁻⁰.⁰⁰⁷ˣ = 0.79
Logarithm both sides with natural logarithm:
ln(e⁻⁰.⁰⁰⁷ˣ) = ln(0.79)
-0.007x ln(e) = -0.24
-0.007x 1 = -0.24
-0.007x = -0.24
x = -0.24 / -0.007
x ≈ 35
