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?
You will need Newton's Law of Cooling formula, which is
T(t) = Tm + (T0 - Tm)e^(kt)
where T(t) is the current temperature of the object
Tm is the surrounding temperature and
T0is the starting temperature
t is in minutes, and k is a constant
first we have to find the value of k
75 = 25 + (90-25)e^(10k)
50 = 65 e^(10k)
.76923 = e^(10k)
10k = ln .76923
k = -.026236
so now
50 = 25 + (90-25)e^(-.026236t)
25/65 = e^(-.026236t)
(-.026236t) = ln(25/65)
t = 36.42 minutes
Thank you!!!
To determine when the temperature of the object will be 50 degrees Celsius, we can use Newton's Law of Cooling. The law states that the rate of change of temperature of an object is directly proportional to the difference between its own temperature and the surrounding temperature.
Step 1: Calculate the cooling constant (k)
The cooling constant (k) can be calculated using the formula:
k = (T_initial - T_surrounding) / (T_final - T_surrounding)
Where:
T_initial = Initial temperature of the object
T_surrounding = Temperature of the surrounding (room temperature)
T_final = Final temperature of the object
In this case:
T_initial = 90 degrees Celsius
T_surrounding = 25 degrees Celsius
T_final = 75 degrees Celsius
k = (90 - 25) / (75 - 25)
k = 65 / 50
k = 1.3
Step 2: Use the equation for Newton's Law of Cooling
The equation for Newton's Law of Cooling is:
T(t) = T_surrounding + (T_initial - T_surrounding) * e^(-k * t)
Where:
T(t) = Temperature of the object at time t
T_surrounding = Temperature of the surrounding (room temperature)
T_initial = Initial temperature of the object
k = Cooling constant
t = Time
We need to solve for t when T(t) = 50 degrees Celsius.
50 = 25 + (90 - 25) * e^(-1.3 * t)
Step 3: Solve the equation for t
To find t, we need to solve the equation for t using logarithms. After rearranging the equation, it becomes:
e^(-1.3 * t) = (50 - 25) / (90 - 25)
e^(-1.3 * t) = 25 / 65
e^(-1.3 * t) = 5 / 13
Take the natural logarithm of both sides:
ln(e^(-1.3 * t)) = ln(5 / 13)
-1.3 * t = ln(5 / 13)
Solve for t:
t = ln(5 / 13) / -1.3
Using a calculator, we find:
t ≈ 23.366 minutes
Therefore, the temperature of the object will be approximately 50 degrees Celsius after 23.366 minutes.
To determine when the object's temperature will reach 50 degrees Celsius, we can use Newton's law of cooling. According to Newton's law of cooling, the rate of change of temperature of an object is directly proportional to the difference between its temperature and the ambient (room) temperature.
Let's set up an equation based on this law.
The rate of change of temperature can be expressed as:
dT/dt = -k(T - T_a)
Where:
- dT/dt represents the rate of change of temperature with respect to time
- k is the cooling constant (which depends on factors like the object's properties and the surrounding medium)
- T is the temperature of the object
- T_a is the ambient (room) temperature
Since we know that:
- The initial temperature of the object is 90 degrees Celsius
- The ambient temperature is 25 degrees Celsius
- The temperature of the object after 10 minutes is 75 degrees Celsius
We can substitute these values into the equation:
- k(90 - 25) = -k(75 - 25)
Simplifying further:
65k = 50k
Now we can solve for k:
65k - 50k = 0
15k = 0
k = 0
It appears that k is equal to zero in this case, which means the rate of change of temperature is zero. However, in reality, this is not possible. It is likely that there is an error or missing information in the given data.