Calculating Energy and Mass in Espresso Machine and Analyzing Cooling of Coffee

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In the discussion, participants calculate the energy required to heat milk in an espresso machine and analyze the cooling of coffee. The energy needed to heat 150 g of milk from 20 °C to 80 °C is determined to be 36,000 J, and the mass of steam condensed is initially miscalculated but later clarified to be approximately 16.4 g. The cooling of coffee is debated, with one participant suggesting the cooling rate is not exponential, and another estimating the coffee's temperature after 12 minutes to be around 29 °C. The conversation also addresses the heat transfer involved when steam condenses into water and the subsequent temperature changes, emphasizing the importance of correctly applying the principles of thermodynamics. Overall, the discussion highlights the complexities of heat transfer calculations in practical scenarios.
  • #31
TSny said:
In the video, the numbers were chosen to make the percent change exactly 50% for two consecutive data points. For real data, you cannot expect the data to be perfectly exponential, or perfectly linear, etc. You are just trying to discover if the data is "well described" by exponential behavior, or linear behavior, etc.

For the data in the problem, the percent decrease is fairly constant at around 15%. If you had to pick between exponential or linear behavior, which would you choose?

For determining the temperature at t = 12 s, I would not recommend your method.
In that case it indeed can be described by exponential. I though the numbers should be exactly the same, and not somewhat similar like in my case.

Do you have a suggestion how to find the temperature at t = 12 s?
 
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  • #32
moenste said:
Do you have a suggestion how to find the temperature at t = 12 s?
First get the temperature at t = 10 seconds using an assumption of exponential behavior. Then go to t = 12 s.
 
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  • #33
TSny said:
First get the temperature at t = 10 seconds using an assumption of exponential behavior. Then go to t = 12 s.
But what percentage to use?

48 °C / 41 °C = 1.17
41 °C / t10 s = 1.17
t10 s = 35.02 °C

35.02 °C / t12 s = 1.17
t12 s = 29.93 °C

Like that?
 
  • #34
You could average the four percent changes that you initially calculated. Use that average percent change to fill out the rest of the table.
 
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