Suppose you have a hot cup of coffee in a room where the temp is 45 Celcius. Let y(t) represent the temp. of coffee as a function of the number of minutes t that have passed since the coffee was poured
a) write a differential equation that applies to newtons law of cooling. use k as constant of proportionality.
the heat flow is proportional to the difference in temperature.
dT/dt= -k(T-Tair) T air is a constant,
after this, i am to solve the differential but I am confused as to what I would make T air.
dT/dt=-k(45-Tair) what do i do with the other 2 variables since k is a constant of proportionality. and T air is a constant
Nevermind I didn't read the question correctly. I got it! Thank You!
To write a differential equation that applies to Newton's Law of Cooling for the given scenario, we start by noting that Newton's Law of Cooling states that the rate of change of temperature of an object is directly proportional to the difference between its temperature and the ambient temperature.
Let's denote the ambient temperature as T_a and the initial temperature of the coffee as T_0. In this case, T_a is the room temperature of 45 degrees Celsius, and we want to find the function y(t) that represents the temperature of the coffee as a function of time t.
The differential equation can be expressed as:
dy/dt = -k(y - T_a)
Here, dy/dt represents the rate of change of temperature, and k is the constant of proportionality.
The term (y - T_a) represents the difference between the current temperature of the coffee (y) and the ambient temperature (T_a). According to Newton's Law of Cooling, the rate of change of temperature is directly proportional to this difference.
The negative sign (-) indicates that the temperature of the coffee is decreasing when it is higher than the ambient temperature (y > T_a), and it is increasing when it is lower than the ambient temperature (y < T_a).
Therefore, the given differential equation captures the relationship between the rate of change of temperature and the difference between the coffee temperature and the ambient temperature.