4. When a cake is removed from a pot, its temperature is measured at 3000C. Three minutes later its temperature is 200o C. How long will it take for the cake to cool off to a room temperature of 70oC?
are we using linear cooling, or Newton's law?
To determine how long it will take for the cake to cool off to a room temperature of 70°C, we can use the concept of exponential decay.
The equation that models the temperature of the cake over time is given by:
T(t) = Ta + (Ti - Ta) * exp(-kt)
Where:
T(t) is the temperature of the cake at time t
Ta is the ambient temperature (room temperature) of 70°C
Ti is the initial temperature of the cake, which is 300°C
k is the cooling constant
t is the time in minutes
To find the cooling constant, we can use the given information. We know that after three minutes, the temperature of the cake is 200°C. So we can substitute these values into the equation:
200 = 70 + (300 - 70) * exp(-3k)
Simplifying the equation:
130 = 230 * exp(-3k)
To solve for k, we can take the natural logarithm of both sides:
ln(130/230) = -3k
k = -ln(130/230)/3
Once we have the value of k, we can substitute it back into the equation to find the time it takes for the cake to cool off to 70°C.
70 = 70 + (300 - 70) * exp(-(k)(t))
Simplifying the equation:
0 = 230 * exp(-(k)(t))
Now, we can solve for t by taking the natural logarithm of both sides:
ln(0) = -kt
t = -ln(0)/k
Since the natural logarithm of 0 is undefined, it means that the cake will never cool off to a room temperature of 70°C based on the given information.