Early in the morning the heater of house breaks down. The outside temperature is −7.0ºC. As a result, the inside temperature drops from 22 to 18ºC in 45 min. How much longer will it take for the inside temperature to fall by another 4.0 Cº? Assume that the outside temperature does not change and that Newton’s law of cooling applies; see problem 1.

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:

Approved