The rate that an object cools is directly proportional to the difference between its temperature ( in Kelvins) at that time and the surrounding temperature (in Kelvins). If an object is initially at 35K, and the surrounding temperature remains constant at 10K, it takes 5 minutes for the object to cool to 25K. How long will it take for the object to cool to 20K?
You are assuming Newton's law of cooling. This leads to a differential equation of T(t) that has a simple exponential decay law. I will skip these steps.
Let k be the cooling rate constant and T' be the surrounding temperature (10 K in this case). t is time in minutes.
The temperature decay equation is
T - T' = 25 e^(-kt)
where the 25 K is the initial temperature difference relative to the surroundings.
Note that dT/dt = -k*(T-T'), as required by the cooling law, and that T(t=0) = 25 + T'
15 = 25*e^(-5k)
-5k = ln(0.6)
k= 0.102 min^-1
When T = 20K,
10 = 25*e^(-0.102 t)
0.4 = e^(-0.102 t)
-.102t = ln(0.4)
t = 9.0 minutes
Well, considering the object is currently at 25K and it took 5 minutes to cool down from 35K to 25K, we can calculate that the rate of cooling is 2K per minute.
Now, if we want to find out how long it will take for the object to cool to 20K, we can simply subtract 20K from 25K and divide that by the rate of cooling, which is 2K per minute.
So, (25K - 20K) / 2K per minute gives us 5 minutes.
Therefore, it will take 5 minutes for the object to cool to 20K. Just remember, don't rush the object, it's trying to chill.
To find out how long it will take for the object to cool to 20K, we can use the same relationship between cooling rate and temperature difference.
Let's denote the time it takes for the object to cool to 20K as t minutes.
We know that the initial temperature of the object is 35K and the surrounding temperature is 10K. So, initially, the temperature difference is 35K - 10K = 25K.
According to the given information, it takes 5 minutes for the object to cool from 35K to 25K. Therefore, in 5 minutes, the temperature difference decreases by 25K - 20K = 5K.
Since the cooling rate is directly proportional to the temperature difference, we can set up a proportion:
(5 minutes) / (5 Kelvin) = t minutes / (25 Kelvin - 20 Kelvin)
Simplifying the proportion:
1 minute = t / 5
Cross multiplying:
t = 5 minutes
Therefore, it will take 5 minutes for the object to cool from 25K to 20K.
To solve this problem, we need to use the concept of exponential decay, which describes the cooling process of an object. We can represent the temperature of the object at any given time t using the equation:
T(t) = T0 + (Ts - T0) * e^(-kt),
where T(t) is the temperature of the object at time t, T0 is the initial temperature of the object, Ts is the surrounding temperature, e is the mathematical constant (approximately equal to 2.71828), and k is the constant of proportionality.
In this case, the initial temperature of the object (T0) is 35K, the surrounding temperature (Ts) is 10K, and we want to find the time it takes for the object to cool to 20K.
We can start by using the given information to find the value of the constant of proportionality (k):
T(5) = 25
35 + (10 - 35) * e^(-5k) = 25
-25 * e^(-5k) = -10
e^(-5k) = 10/25
e^(-5k) = 2/5
Now, take the natural logarithm (ln) of both sides of the equation to solve for k:
ln(e^(-5k)) = ln(2/5)
-5k * ln(e) = ln(2/5)
-5k = ln(2/5)
Next, solve for k:
k = ln(2/5) / -5 ≈ 0.2231
Now that we have the value of k, we can use the equation to calculate the time it takes to cool to 20K:
T(t) = T0 + (Ts - T0) * e^(-kt)
20 = 35 + (10 - 35) * e^(-0.2231 * t)
20 - 35 = -25 * e^(-0.2231 * t)
-15 = -25 * e^(-0.2231 * t)
Divide both sides of the equation by -25:
(-15)/(-25) = e^(-0.2231 * t)
0.6 = e^(-0.2231 * t)
Now, take the natural logarithm of both sides:
ln(0.6) = -0.2231 * t
Finally, solve for t:
t = ln(0.6) / -0.2231
Using a calculator, we can find that t ≈ 6.777.
Therefore, it will take approximately 6.777 minutes for the object to cool to 20K.