Newton's Law of Cooling is a principle that describes how the temperature of an object changes as it interacts with its surroundings. According to this law, the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings.
i) The temperature of the water after 20 minutes is approximately 73.47°C
ii) The time at which the temperature of the water is 60°C is approximately 31.83 minutes.
Newton's Law of Cooling is a principle that describes how the temperature of an object changes as it interacts with its surroundings. According to this law, the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings.
To solve the given problem, we can use Newton's Law of Cooling equation:
dT/dt = -k(T - Ts)
where dT/dt represents the rate of change of temperature with respect to time, T represents the temperature of the object, Ts represents the temperature of the surroundings, and k is a constant that depends on the object's characteristics and the environment.
Now let's solve the problem step by step:
i) To find the temperature of the water after 20 minutes, we need to determine the value of T at that time.
We are given that the initial temperature of the water (T0) is 100°C and it cools to 88°C in 10 minutes. This gives us two data points: (0, 100) and (10, 88).
Using these data points, we can find the value of k by rearranging the equation and substituting the values:
k = (ln((T - Ts)/(T0 - Ts))) / t
k = (ln((88 - 25)/(100 - 25))) / 10
k ≈ -0.0543
Now we can substitute the values of k, T0, Ts, and t into the equation to find the temperature at 20 minutes:
T = Ts + (T0 - Ts) * exp(-k * t)
T = 25 + (100 - 25) * exp(-(-0.0543) * 20)
T ≈ 73.47°C
Therefore, the temperature of the water after 20 minutes is approximately 73.47°C.
ii) To find the time at which the temperature of the water is 60°C, we need to determine the value of t at that temperature.
Using the equation, we can rearrange it to solve for t:
t = (ln((T - Ts)/(T0 - Ts))) / (-k)
t = (ln((60 - 25)/(100 - 25))) / (-0.0543)
t ≈ 31.83 minutes
Therefore, the time at which the temperature of the water is 60°C is approximately 31.83 minutes.