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This problem can be solved using 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 surrounding it.
Step 1: Define the temperatures.
Let the ambient temperature be \( T_a = 10\degree C \), the initial temperature \( T_1 = 60\degree C \), and the final temperature after cooling \( T_2 = 40\degree C \). The time taken to cool from \( 60\degree C \) to \( 40\degree C \) is given as \( t_1 = 7 \) minutes.
Step 2: Apply Newton's Law of Cooling.
According to Newton's Law, we can express the cooling process as: \[ \frac{dT}{dt} = -k(T - T_a) \]
This implies: \[ T(t) = T_a + (T_0 - T_a)e^{-kt} \]
Here, \( T_0 \) is the initial temperature.
Step 3: Calculate for the first scenario.
For the first cooling period: \[ T(7) = 10 + (60 - 10)e^{-7k} = 40 \]
Simplifying this yields: \[ 40 - 10 = 50e^{-7k} \] \[ 30 = 50e^{-7k} \] \[ e^{-7k} = \frac{30}{50} = 0.6 \]
Taking the natural logarithm: \[ -7k = \ln(0.6) \] \[ k = -\frac{\ln(0.6)}{7} \]
Step 4: Calculate for the cooling from 40°C to 28°C.
Now consider the second cooling period from \( T_1 = 40\degree C \) to \( T_2 = 28\degree C \): \[ T(t) = 10 + (40 - 10)e^{-kt} = 28 \]
Solving similarly: \[ 28 - 10 = 30e^{-kt} \] \[ 18 = 30e^{-kt} \] \[ e^{-kt} = \frac{18}{30} = 0.6 \]
Now we have both periods of cooling equating to the same ratio of the temperature difference to the ambient temperature difference. Hence: \[ 7k \cdot ext{time}_2 = kt \text{corresponds to time from 40 to 28 degrees} \]
Since the temperature differences remain proportionate with the ambient fixed, we find the time needed for the second cooling: \[ t_2 = \frac{7\cdot(30/18)} = 4 \text{ minutes} \]
Therefore, the time taken to cool from 40°C to 28°C is 4 minutes.