coffee cup simulation

Q1. How does cooling behavior change if Rth & C change?

Well, Rth is the thermal resistance of coffee... so it makes sense that if the resistance is increased, then it should take longer for the coffee to cool down, and if it is decreased, then it should also speed up down the cooling process. With C being the rate of change of thermal energy vs temperature rise, we should also expect the same relationship- if there is a positive change, then it should slow the cooling process since there is energy being added or increased in someway, while if there is a negative change, it should speed up the cooling process since energy is being lost.

Looking at the equation

dT= ((∆t)/(Rth*C))dt

We can see that Rth and C are in the denominator, thus if either of those values are increased, dt will be increased.

Q2. Adding a heater: What is a good P value if we want our room temp coffee to heat up to the Starbucks ideal 84 degrees?

This situation is governed (well, modeled) by the equation...

dT = dE/C = [ (P/C)-((T-Tair)/(Rth*C)) ] dt

 Since we know that Tair = 293K, and our ideal temperature is 357K, dT/dt will be = 357-293 = 64. Now, solving for P, we get a value of 75.36. Plugging this into our heat simulation program in MATLAB, we get this plot...

We can see that it levels off in the 350-360 range. Indeed, printing the final temperature value in MATLAB gave us 356... close enough (error due to rounding, probably.)