This exercise uses Newton's Law of Cooling.
A kettle full of water is brought to a boil in a room with temperature 20°C. After 15 minutes the temperature of the water has decreased from 100° to 75°C. Find the temperature after another 13 min. (Round your answer to one decimal place.)
Well, isn't the kettle just boiling with excitement? Let's use Newton's Law of Cooling to figure this out!
According to Newton's Law of Cooling, the rate at which a hot object cools is directly proportional to the temperature difference between the object and its surroundings. In this case, the surroundings are at a constant temperature of 20°C.
Now, let's calculate the cooling constant, shall we? We're told that the water cools from 100°C to 75°C in 15 minutes. So, the temperature difference is 100 - 20 = 80°C.
Using Newton's Law of Cooling, we can write the equation as:
ΔT/Δt = -k(T - Ts)
Where ΔT is the temperature difference, Δt is the time difference, k is the cooling constant, T is the temperature of the water at any given time, and Ts is the temperature of the surroundings.
Plugging in the values we have:
-80/15 = -k(100 - 20)
Simplifying:
-80/15 = -80k
Solving for k:
k = 80/15 ≈ 5.33
Alright, we have our cooling constant! Now let's find the temperature after another 13 minutes.
Using the same equation, but plugging in T = 75°C (because it cooled to that) and Δt = 13:
-ΔT/Δt = k(T - Ts)
-ΔT/13 = 5.33(75 - 20)
Solving for ΔT:
-ΔT/13 = 5.33(55)
ΔT ≈ -401.15
Since we're looking for the temperature after 13 minutes, we add this change to the current temperature:
75 + (-401.15) ≈ -326.15°C
But hold on a second, negative temperatures? That's a bit chilly, don't you think? Let's put it in terms of the original scale.
Converting it back, we get:
-326.15°C ≈ -787.07°F
So, the temperature after another 13 minutes is approximately -787.07°F. Mind if I join you for a cup of boiling laughter instead?
To solve this problem, we can use Newton's Law of Cooling, which states that the rate of cooling of an object is directly proportional to the temperature difference between the object and its surroundings.
Let's denote the initial temperature of the water as T0, the final temperature as T, the time at which the final temperature is reached as t, and the room temperature as Tr.
According to Newton's Law of Cooling, we have the following differential equation:
dT/dt = -k(T - Tr),
where k is the cooling constant that depends on the properties of the water and the kettle.
Given that the initial temperature (T0) is 100°C and the temperature after 15 minutes (t = 15 min) is 75°C, we can set up the following equation:
dT/dt = -k(T - Tr),
75 = 100 - k(100 - 20).
Simplifying the equation, we get:
25 = k * 80.
Dividing both sides by 80, we find:
k = 25/80.
To find the temperature after another 13 minutes (t = 13 min), we can solve the differential equation using an initial value problem:
dT/dt = -k(T - Tr),
with the initial condition T(15) = 75.
Integrating both sides of the equation, we get:
∫dT/(T - Tr) = -∫k dt,
ln|T - Tr| = -kt + C,
where C is the constant of integration.
Applying the initial condition T(15) = 75, we have:
ln|75 - 20| = -k * 15 + C,
ln|55| = -k * 15 + C.
Simplifying, we get:
ln(55) = -15 * (25/80) + C,
ln(55) = -375/80 + C.
Calculating the value of C, we find:
C = ln(55) + 375/80.
Substituting the value of C back into the equation, we have:
ln|T - 20| = -kt + ln(55) + 375/80.
Simplifying further, we get:
ln|T - 20| = -kt + ln(55) + 375/80.
Using the given time t = 13 min, we can solve for the temperature T after another 13 minutes:
ln|T - 20| = -13 * (25/80) + ln(55) + 375/80.
Simplifying, we find:
ln|T - 20| = -325/80 + ln(55) + 375/80.
Calculating the right-hand side of the equation, we have:
ln|T - 20| = 1217/80.
Now, exponential both sides of the equation:
|T - 20| = e^(1217/80).
Since the temperature cannot be negative, we take the positive value of the absolute value:
T - 20 = e^(1217/80).
Solving for T, we get:
T = e^(1217/80) + 20.
Now, we can calculate the temperature after another 13 minutes:
T = e^(1217/80) + 20 ≈ 87.8°C.
Therefore, the temperature after another 13 minutes is approximately 87.8°C.
To find the temperature of the water after another 13 minutes, we can use Newton's Law of Cooling, which states:
T(t) = T0 + (T1 - T0) * e^(-kt)
Where:
- T(t) is the temperature at time t
- T0 is the initial temperature
- T1 is the temperature after a certain time
- k is the cooling constant
- e is the base of the natural logarithm (approximately 2.71828)
In this case, T0 is 100°C (initial temperature), T1 is 75°C (temperature after 15 minutes), and we want to find the temperature after another 13 minutes. Let's calculate the cooling constant, k, first.
Using the information given, we can rearrange the equation to solve for k:
k = ln((T1 - T0) / (T(t) - T0)) / t
where ln represents the natural logarithm and t is the time interval (in this case, 15 minutes).
Plugging in the values:
k = ln((75 - 100) / (T(t) - 100)) / 15
Now, we can solve for the temperature T(t) after another 13 minutes by plugging in the values into the original equation:
T(t) = 100 + (75 - 100) * e^(-k * 13)
After evaluating this equation, rounding the answer to one decimal place, we will obtain the temperature after another 13 minutes.
T(t) = 20 + (100-20)e^(kt)
using T(15) = 75, we have
20 + 80e^(15k) = 75
k = -0.025
So now find T(15+13)