Final answer:
To find the period of oscillations, calculate the time it takes for temperature to change from rising to falling or vice versa. The temperature drops at a rate of 2 K per minute when the heater is off, and rises at a rate of 4 K per minute when the heater is on. The period of oscillations is the sum of the time it takes to reach the neutral zone from falling to rising and vice versa.
Step-by-step explanation:
To find the period of oscillations, we need to calculate the time it takes for the temperature to change from rising to falling, or vice versa. From the given information, when the heater is off, the temperature drops at a rate of 2 K per minute, and when the heater is on, the temperature rises at a rate of 4 K per minute. The period of oscillations can be calculated by finding the time it takes for the temperature to change from falling to rising or vice versa.
When the heater is off, the temperature decreases by 2 K per minute. To reach the neutral zone at +/−6.46 K, it will take 3.23 minutes (6.46 K / 2 K per min = 3.23 min). Similarly, when the heater is on, the temperature increases by 4 K per minute. To reach the neutral zone, it will take 1.61 minutes (6.46 K / 4 K per min = 1.61 min). Therefore, the period of oscillations is the sum of these two times, which is 4.84 minutes (3.23 min + 1.61 min = 4.84 min).
To plot the water temperature versus time, we can divide the time into intervals of the period of oscillation (4.84 minutes). At each interval, the temperature will be either rising (4 K per min) or falling (2 K per min). We can start with an initial temperature at or near the set-point (323 K) and calculate the temperature at each interval until we cover the desired time range.