Answer :
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⁻ⁿ*¹⁰
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⁻⁰.⁰⁰⁷*ˣ
180 - 68 = 142 * e⁻⁰.⁰⁰⁷*ˣ
112 = 142 * e⁻⁰.⁰⁰⁷⁾*ˣ
e⁻⁰.⁰⁰⁷*ˣ = 112/142
e⁻⁰.⁰⁰⁷*ˣ = 0.79
Logarithm both sides with natural logarithm:
ln(e⁻⁰.⁰⁰⁷*ˣ) = 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
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⁻ⁿ*¹⁰
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⁻⁰.⁰⁰⁷*ˣ
180 - 68 = 142 * e⁻⁰.⁰⁰⁷*ˣ
112 = 142 * e⁻⁰.⁰⁰⁷⁾*ˣ
e⁻⁰.⁰⁰⁷*ˣ = 112/142
e⁻⁰.⁰⁰⁷*ˣ = 0.79
Logarithm both sides with natural logarithm:
ln(e⁻⁰.⁰⁰⁷*ˣ) = 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
Final answer:
Using the cooling rate obtained from the given data (1 degree per minute), it will take approximately 20 more minutes for the coffee to cool from 200 degrees Fahrenheit to 180 degrees Fahrenheit.
Explanation:
You asked about how long it will take for a cup of coffee that initially was at 210 degrees Fahrenheit to cool to 180 degrees Fahrenheit in a room that is 68 degrees Fahrenheit, given that it cooled from 210 degrees to 200 degrees Fahrenheit in 10 minutes. Using Newton's law of cooling, we can approximate the cooling rate of the coffee. Newton's law of cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (i.e., the temperature of its surroundings).
To solve this problem, we would typically use the formula derived from Newton's law, which is a first-order differential equation. However, as we have been provided with a specific cooling period (10 minutes for a 10 degree Fahrenheit drop), we can estimate the total cooling time by setting up a ratio, assuming that the rate of cooling is roughly constant over these temperature intervals, which is a common simplification for small temperature ranges.
The coffee cooled 10 degrees in 10 minutes, so it cools at a rate of 1 degree per minute. If the coffee is currently at 200 degrees Fahrenheit and needs to reach 180 degrees Fahrenheit, it has 20 more degrees to cool. Assuming a constant cooling rate:
Time to cool from 200 to 180 degrees = 20 degrees * 1 minute/degree = 20 minutes
Therefore, it will take approximately 20 more minutes for the coffee to reach 180 degrees Fahrenheit.
