An object is heated to 90 degrees Celcius and is then allowed to cool in a room whose air temperature is 25 degrees celcius. If the temperature of the object is 75 degrees celcius after 10 minutes, when will its temperature be 50 degrees celcius?
I have no idea how to do this. There are just to many numbers involved!!
Don't worry! Let me guide you through this problem. We can use Newton's Law of Cooling, which states:
dT/dt = -k(T - T_env)
Where:
- dT/dt is the rate of change in the object's temperature with respect to time,
- k is a cooling constant (depends on the object and medium, but remains constant for this problem),
- T is the object's temperature at a given time,
- T_env is the temperature of the environment (in this case, room air temperature).
First, we'll find the cooling constant k. To do that, we can use the information that the object's temperature is 75° C after 10 minutes (10 minutes = 600 seconds).
We know that:
T = 75° C
T_env = 25° C
dT/dt = (75 - 90) / 600 = -15 / 600 = -1/40 °C/s
Now, we can set up the equation:
-1/40 = -k(75 - 25)
Solve for k:
k = (1/40) / 50 = 1/2000 (unit doesn't matter here as it will cancel out)
Now that we have the cooling constant (k), we can find the time it takes for the object to reach 50° C.
Let's call that time t_seconds. Now we need to set up the equation with the information we have:
dT/dt = -k(T - T_env)
We have:
T = 50° C
T_env = 25° C
k = 1/2000
Now we'll find dT/dt when the temperature of the object is 50° C:
dT/dt = (50 - 90) / t_seconds
Inserting the values into the equation:
(50 - 90) / t_seconds = -(1/2000) * (50 - 25)
Solving for t_seconds:
-40 / t_seconds = -1/80
t_seconds = 40 * 80 = 3200 seconds
Now convert the seconds to minutes:
3200 seconds = 3200 / 60 = 53.33 minutes
So, it will take approximately 53.33 minutes for the object's temperature to reach 50° C.
To solve this problem, we can use Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature.
Let's denote:
- T(t) as the temperature of the object at time t
- T_a as the ambient temperature (25 degrees Celsius in this case)
- k as the cooling constant
Using the given information, we can set up an equation:
dT/dt = -k(T - T_a)
We know that at t = 0 (initially), T(0) = 90 degrees Celsius.
And at t = 10 minutes, T(10) = 75 degrees Celsius.
Now, let's solve for the cooling constant (k) using these initial and final temperature values.
dT/dt = -k(T - T_a)
(75 - 25)/(10 - 0) = -k(75 - 25)
50/10 = -k(50)
5 = -50k
k = -5/50
k = -1/10
Now, we have the cooling constant (k = -1/10).
To find the time (t) when the object's temperature is 50 degrees Celsius, we can set up the equation:
dT/dt = -k(T - T_a)
T(t) - T_a = (T(0) - T_a) * e^(-kt)
Substituting the known values:
50 - 25 = (90 - 25) * e^(-(-1/10)t)
Simplifying the equation:
25 = 65 * e^(1/10t)
Dividing both sides by 65:
25/65 = e^(1/10t)
Taking the natural logarithm of both sides:
ln(25/65) = 1/10t
Simplifying:
t = 10ln(25/65)
Using a calculator, we find that t is approximately 9.82 minutes.
Therefore, the object's temperature will be 50 degrees Celsius after approximately 9.82 minutes.
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.
First, let's define the variables:
T0 = Initial temperature of the object (90 degrees Celsius)
Ta = Air temperature (25 degrees Celsius)
T = Temperature of the object at any given time
t = Time in minutes
Using the given information, we can write the equation for Newton's Law of Cooling as:
dT/dt = -k(T - Ta)
Where dT/dt represents the rate of change of temperature with respect to time, and k is the cooling constant.
Now, let's solve for the cooling constant, k, using the given data:
dT/dt = (75 - 25) / 10
dT/dt = 5 degrees Celsius per minute
So, we have the equation:
5 = -k(75 - 25)
Simplifying, we get:
5 = -50k
Solving for k, we find:
k = -1/10
Now, let's find the time it takes for the object to cool to 50 degrees Celsius. We'll use the equation:
dT/dt = -k(T - Ta)
Let T = 50 and Ta = 25:
5 = -(-1/10)(50 - 25)
Simplifying, we get:
5 = -(-1/10)(25)
Solving for t, we find:
t = 25 minutes
So, the object will reach a temperature of 50 degrees Celsius after 25 minutes.