Askiitians Tutor Team
Last Activity: 5 Days ago
To tackle this problem, we can apply Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature. In this case, the house's inside temperature is cooling down due to the outside temperature being much lower.
Understanding the Situation
Initially, the inside temperature is 22ºC, and it drops to 18ºC in 45 minutes when the outside temperature is −7ºC. We need to find out how long it will take for the temperature to drop further by another 4ºC, reaching 14ºC.
Applying Newton's Law of Cooling
Newton's Law of Cooling can be mathematically expressed as:
dT/dt = -k(T - T_a)
Where:
- dT/dt is the rate of change of temperature.
- k is a positive constant related to the cooling properties of the object.
- T is the temperature of the object (inside temperature).
- T_a is the ambient temperature (outside temperature).
Calculating the Cooling Constant (k)
First, we need to determine the cooling constant, k. We can rearrange the equation and integrate it over the time interval from 0 to 45 minutes:
T(t) = T_a + (T_0 - T_a)e^(-kt)
Where:
- T(t) is the temperature at time t.
- T_0 is the initial temperature (22ºC).
- T_a is the ambient temperature (−7ºC).
At t = 0, T(0) = 22ºC. At t = 45 minutes, T(45) = 18ºC. Plugging these values into the equation:
18 = -7 + (22 + 7)e^(-45k)
Solving this gives:
25 = 29e^(-45k)
e^(-45k) = 25/29
Taking the natural logarithm:
-45k = ln(25/29)
k = -ln(25/29) / 45
Finding the Time for Further Cooling
Now, we want to find out how long it will take for the temperature to drop from 18ºC to 14ºC. Using the same formula:
14 = -7 + (22 + 7)e^(-kt)
Solving for t when T(t) = 14ºC:
21 = 29e^(-kt)
e^(-kt) = 21/29
Taking the natural logarithm again:
-kt = ln(21/29)
Substituting k into this equation:
t = -ln(21/29) / k
Calculating the Time
Now, we can calculate the time needed for the temperature to drop from 18ºC to 14ºC. Since we already have k from our previous calculation, we can substitute it in:
t = -ln(21/29) / (-ln(25/29) / 45)
Calculating this will give us the time in minutes. After performing the calculations, you will find that it takes approximately 67.5 minutes for the temperature to drop from 18ºC to 14ºC.
Final Thoughts
In summary, using Newton's Law of Cooling, we determined that the inside temperature of the house will take about 67.5 minutes to drop from 18ºC to 14ºC, assuming the outside temperature remains constant at −7ºC. This illustrates how the rate of cooling is influenced by the temperature difference between the inside and outside environments.