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

In summary: C?The specific latent heat of steam is 2.2 MJ/kg, which means that in order for steam to condense into water at 100 degrees, 2.2 MJ of heat needs to be removed from 1 kg of steam. In this case, we have 0.1 kg of steam, so the total heat that needs to be removed is 0.1 * 2.2 MJ = 220 kJ. Combining this with the 42 J of heat needed to change the temperature from 100 oC to 90 oC, we get a total of 220.042 kJ of heat that needs to be removed from the steam in order for it to
  • #1
moenste
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Homework Statement


(a) In an espresso coffee machine, steam at 100 °C is passed into milk to heat it. Calculate (i) the energy required to heat 150 g of milk from room temperature (20 °C) to 80 °C, (ii) the mass of steam condensed.

(b) A student measures the temperature of the hot coffee as it cools. The results are given below:
d22f1b6d6801.jpg

A friend suggests that the rate of cooling is exponential. (i) Show quantitatively whether this suggestion is valid. (ii) Estimate the temperature of the coffee after a total of 12 min.
  • Specific heat capacity of milk = 4.0 kJ kg-1 K-1
  • Specific heat capacity of water = 4.2 kJ kg-1 K-1
  • Specific latent heat of steam = 2.2 MJ kg-1
Answers: (a) (i) 3.6 * 104 J, (ii) 15.8 g

2. The attempt at a solution
(a) (i) ΔQ = mcΔθ = 0.15 kg * 4000 J kg-1 K-1 * (80-20 °C) = 36 000 J -- the energy required to heat 150 g of milk from room temperature of 20 degrees to 80 degrees.

(a) (ii) ΔQ = ml → to find mass m = ΔQ / l = 36 000 J / 2.2 * 106 J kg-1 = 0.016 kg or 16.4 g. Which is wrong. What did I miss?

(b) (i) Can't say that it's exponential, since the increase is not the same in percentage.

(b) (ii) The temperature should be 29 °C, because we have a difference of two twice and a difference of one once. So, I guess every difference should be two times the number: 2 x 2, 2 x 1, 2 x 0.

1b2376b65c9b.jpg


In sum: what's missing in (a) (ii) and is my logic in (b) (i-ii) correct?
 
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  • #2
Check the temperature change. 80-60 ??
 
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  • #3
CWatters said:
Check the temperature change. 80-60 ??
Oh yes, it's a typo, I guess I was thinking of sixty (80-20 = 60) and that's why got 80-60. The answer is correct though in (a) (i).
 
  • #4
moenste said:
(a) (ii) ΔQ = ml → to find mass m = ΔQ / l = 36 000 J / 2.2 * 106 J kg-1 = 0.016 kg or 16.4 g. Which is wrong. What did I miss?
Any ideas please?
 
  • #5
When the steam condenses, what is the temperature of the resultant water immediately after the condensation? What happens subsequently to the temperature of this water as it comes to thermal equilibrium with the milk?
 
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  • #6
TSny said:
When the steam condenses, what is the temperature of the resultant water immediately after the condensation? What happens subsequently to the temperature of this water as it comes to thermal equilibrium with the milk?
Sorry, what water are you talking about? I think we only have milk in part (a).
 
  • #7
moenste said:
Sorry, what water are you talking about?
What happens to the steam when it is added to the milk?
 
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  • #8
TSny said:
What happens to the steam when it is added to the milk?
The steam heats the milk to 80 degrees. It just heats the milk, as I see it.
 
  • #9
The steam is originally at 100 oC when it is added to the milk. At this temperature, what happens to the steam if some heat is transferred from the steam to the milk?
 
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  • #10
TSny said:
The steam is originally at 100 oC when it is added to the milk. At this temperature, what happens to the steam if some heat is transferred from the steam to the milk?
I would guess it would have some drops or water. The steam would create humidity.
 
  • #11
TSny said:
The steam is originally at 100 oC when it is added to the milk. At this temperature, what happens to the steam if some heat is transferred from the steam to the milk?
I think you mean that when we heat milk to 80 degrees using steam, some water is added to milk. That's why we need to calculate the combined mass of water and milk.

So we have ΔQ = mcΔθ where we need to find the mass of water, m = ΔQ / cΔθ = 36 000 / 4200 * 60 = 0.14 kg is the mass of water.

Then we combine 16.4 g with 142.9 g = 159.3 g. The mass of steam condensed.
 
  • #12
moenste said:
I think you mean that when we heat milk to 80 degrees using steam, some water is added to milk. That's why we need to calculate the combined mass of water and milk.

So we have ΔQ = mcΔθ where we need to find the mass of water, m = ΔQ / cΔθ = 36 000 / 4200 * 60 = 0.14 kg is the mass of water.

Then we combine 16.4 g with 142.9 g = 159.3 g. The mass of steam condensed.
This is not correct.

Try the following exercise. How much heat would need to be removed from 100 g of steam at 100 oC to convert it all to water at 90 oC?
 
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  • #13
TSny said:
This is not correct.

Try the following exercise. How much heat would need to be removed from 100 g of steam at 100 oC to convert it all to water at 90 oC?
m = 0.1 kg
TInitial = 100 °C
TFinal = 90 °C
TLost = -10 °C
c of water = 4200 J / kg K

ΔQ = mcΔθ = 0.1 * 4200 * -10 = -42 J.
 
  • #14
moenste said:
m = 0.1 kg
TInitial = 100 °C
TFinal = 90 °C
TLost = -10 °C
c of water = 4200 J / kg K

ΔQ = mcΔθ = 0.1 * 4200 * -10 = -42 J.
You've correctly calculated the heat required to change the temperature of 100 g of water from 100 oC to 90 oC, except your final number does not have the decimal in the right place.

But this is only part of the answer.
How much heat must first be removed from the steam in order for it to condense into water at 100 oC?
 
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  • #15
TSny said:
You've correctly calculated the heat required to change the temperature of 100 g of water from 100 oC to 90 oC, except your final number does not have the decimal in the right place.

But this is only part of the answer.
How much heat must first be removed from the steam in order for it to condense into water at 100 oC?
90 - 100 = -10 degrees are lost
c of steam at 100 degrees is 1890 J / kg K

ΔQ = 0.1 * 1890 * -10 = -1890 J are lost.
 
  • #16
We have to deal with the steam in two steps:

(1) The steam condenses to water. This is a phase change. So, think about how to calculate heat involved in a phase change. The temperature doesn't change during the phase change.

(2) The water, that used to be steam, now cools to the final temperature.
 
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  • #17
TSny said:
We have to deal with the steam in two steps:

(1) The steam condenses to water. This is a phase change. So, think about how to calculate heat involved in a phase change. The temperature doesn't change during the phase change.

(2) The water, that used to be steam, now cools to the final temperature.
(1) Heat involved in a phase change: ΔQ = ml = 0.1 * 2.2 * 106 = 220 000 J.

(2) ΔQ = mcΔ(θNew - θInitial) = 0.1 * 4200 * -10 = -4200 J
So heat involved 220 000 + (-4200) = 224 200 J

I'm sorry, but this doesn't get me any closer to understanding what is wrong with my original 16.4 g answer.

In your example I don't understand whether the mass is constant? 100 g of steam will become 100 g of water?

I also don't understand what to apply to my problem.

We have got 36 000 J. Do I need to do the same thing but with c = 4200? Then the answer will be 37 800 J. Using this number I get 17.2 g. If I combine 37 800 + 36 000 I'll get 33.5 g. If I subtract I'll get 1800 J and 0.82 g.

I also have for some reason 100 °C. If I use Δθ = 100 - 80 or 100 - 20 it still does not get anywhere: 0.15 * 4200 * 20 = 12 600 J or 0.15 * 4200 * 80 = 50 400 J. None of that get the 15.8 g answer.

If I look for Q = ml = 0.0158 * 2.2 * 106 = 34 760 J. So my Q should be equal to this number. Even if I subract 37 800 - 36 000 I'll get 1800 and then 36 000 - 1800 = 34 200 J and the mass will be m = 15.5 g. Close but still not 15.8 g.
 
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  • #18
moenste said:
(1) Heat involved in a phase change: ΔQ = ml = 0.1 * 2.2 * 106 = 220 000 J.

(2) ΔQ = mcΔ(θNew - θInitial) = 0.1 * 4200 * -10 = -4200 J
So heat involved 220 000 + (-4200) = 224 200 J
OK, good. You've found the total amount of heat given up by 100 g of steam as it goes through the "two-step" process of condensing to water and then cooling to a final temperature of 90 degrees.

In your example I don't understand whether the mass is constant? 100 g of steam will become 100 g of water?
Yes. (Same number of H2O molecules before and after.)

I also don't understand what to apply to my problem.

We have got 36 000 J.
Yes, that's the amount of heat that must be added to the milk to get the final temperature to be 80 degrees. This heat comes from the steam. That steam will start at 100 degrees and then will go through the "two-step"" process to become water at the final temperature.

Let M be the unknown amount of steam required. Can you set up an expression for the total amount of heat given up by this amount of steam? Just follow what you did for the example with 100 grams.
 
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  • #19
TSny said:
Yes. (Same number of H2O molecules before and after.)
Thank you, this helps.

TSny said:
Yes, that's the amount of heat that must be added to the milk to get the final temperature to be 80 degrees. This heat comes from the steam. That steam will start at 100 degrees and then will go through the "two-step"" process to become water at the final temperature.

Let M be the unknown amount of steam required. Can you set up an expression for the total amount of heat given up by this amount of steam? Just follow what you did for the example with 100 grams.
Alright, so we have milk that got it's temperature increased from 20 to 80 and the heat required to do that is equal to 36 000 J.

The steam of mass M was transformed into water of equal mass M and the transaction took ΔQ = ml or Q = (2 200 000 M) Joules. Now we have water that is cooling down from 100 degrees to 80 degrees. ΔQ = mcΔθ or Q = 4200 * (80 - 100) M so Q = -84 000 M.

We have
Q = 2 200 000 M and Q = -84 000 M.

Am I following you correctly?
 
  • #20
moenste said:
We have
Q = 2 200 000 M and Q = -84 000 M.

OK. But when steam condenses to water it loses heat. So, Q for the steam would be negative: Q = -ML, where Q represents heat added to a system.
 
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  • #21
TSny said:
OK. But when steam condenses to water it loses heat. So, Q for the steam would be negative: Q = -ML
You mean Q = -2 200 000 M and Q = -84 000 M?

But here I have positive:
moenste said:
(1) Heat involved in a phase change: ΔQ = ml = 0.1 * 2.2 * 106 = 220 000 J.

(2) ΔQ = mcΔ(θNew - θInitial) = 0.1 * 4200 * -10 = -4200 J
So heat involved 220 000 + (-4200) = 224 200 J
 
  • #22
Yes. I see now that you made two mistakes that canceled their effects. For step 1 you should have gotten -220,000 J for the heat gained. But, you also had

220,000 + (-4200) = 224,200 J, which is incorrect math.

It should have been ΔQ = -220,000 J + (-4200 J) = -224, 200 J. So, the total heat gained by the steam is -224,200 J. Or you could say that the steam gave up 224,200 J of heat.
 
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  • #23
TSny said:
Yes. I see now that you made two mistakes that canceled their effects. For step 1 you should have gotten -220,000 J for the heat gained. But, you also had

220,000 + (-4200) = 224,200 J, which is incorrect math.

It should have been ΔQ = -220,000 J + (-4200 J) = -224, 200 J. So, the total heat gained by the steam is -224,200 J. Or you could say that the steam gave up 224,200 J of heat.
Oh, yes indeed I miscalculated. It should've been 215 800.

But how do we get ΔQ = ml to be negative? We get a positive number and just put a minus sign because we know that this energy was given up and therefore it should be negative? Or it should be like m = - m since we loose 100 g of steam and gain 100 g of water, so Q = ml where m = -0.1 and l = 2 200 000?
 
  • #24
moenste said:
But how do we get ΔQ = ml to be negative? We get a positive number and just put a minus sign because we know that this energy was given up and therefore it should be negative?
Yes. If ΔQ is defined to be the symbol for the heat gained by a system during some process, then ΔQ = ±ML for a substance undergoing a phase change. Choose the + sign if heat is added to the substance during the phase change. Choose the - if heat is given up or "lost" by the substance during the phase change.

For a substance undergoing a temperature change (no phase change), then you can use ΔQ = mcΔT for the heat gained by the substance. The ΔT will take care of the sign automatically.

The signs can be annoying. Sometimes we talk about the amount of heat "lost" by a substance rather than the heat "gained" by a substance. "Heat lost" is the negative of the "heat gained". So, a heat gain of -1000 J is equivalent to a heat loss of +1000 J.

Or it should be like m = - m since we loose 100 g of steam and gain 100 g of water, so Q = ml where m = -0.1 and l = 2 200 000?
I would not recommend this way of thinking about it.
 
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  • #25
TSny said:
Yes. If ΔQ is defined to be the symbol for the heat gained by a system during some process, then ΔQ = ±ML for a substance undergoing a phase change. Choose the + sign if heat is added to the substance during the phase change. Choose the - if heat is given up or "lost" by the substance during the phase change.

For a substance undergoing a temperature change (no phase change), then you can use ΔQ = mcΔT for the heat gained by the substance. The ΔT will take care of the sign automatically.

The signs can be annoying. Sometimes we talk about the amount of heat "lost" by a substance rather than the heat "gained" by a substance. "Heat lost" is the negative of the "heat gained". So, a heat gain of -1000 J is equivalent to a heat loss of +1000 J.I would not recommend this way of thinking about it.
OK, so we have Q = -2 200 000 M and Q = -84 000 M and we have energy lost 2 284 000 M (Joules). How do we use this number to get M? 36 000 / 2 284 000 = M = 15.76 g?
 
  • #26
moenste said:
OK, so we have Q = -2 200 000 M and Q = -84 000 M and we have energy lost 2 284 000 M (Joules). How do we use this number to get M? 36 000 / 2 284 000 = M = 15.76 g?
Yes. That's it.

To summarize, the milk must gain 36 000 J of heat. The heat gained by the steam is (-2 200 000 M) + (-84 000 M) = -2 284 000 M. Or, the steam loses 2 284 000 M Joules of heat. The heat lost by the steam equals the heat gained by the milk:

2 284 000 M = 36 000.

Solving for M gives M = .01576 kg.

If you want to include units in your calculation, then you would need to note that the units for the numbers 2 200 000 and 84 000 are J/kg. So,

(2 284 000 J/kg) * M = 36 000 J
 
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  • #27
TSny said:
Yes. That's it.

To summarize, the milk must gain 36 000 J of heat. The heat gained by the steam is (-2 200 000 M) + (-84 000 M) = -2 284 000 M. Or, the steam loses 2 284 000 M Joules of heat. The heat lost by the steam equals the heat gained by the milk:

2 284 000 M = 36 000.

Solving for M gives M = .01576 kg.

If you want to include units in your calculation, then you would need to note that the units for the numbers 2 200 000 and 84 000 are J/kg. So,

(2 284 000 J/kg) * M = 36 000 J
Thank you, I better understood it with this sentence "The heat lost by the steam equals the heat gained by the milk".

What can be said about (b)? I think it should be correct, at least (b) (i) should.
moenste said:
(b) (i) Can't say that it's exponential, since the increase is not the same in percentage.

(b) (ii) The temperature should be 29 °C, because we have a difference of two twice and a difference of one once. So, I guess every difference should be two times the number: 2 x 2, 2 x 1, 2 x 0.

1b2376b65c9b.jpg
 
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  • #28
Well, as you stated in your first post, the percent increase is fairly consistent at -14% to -15% for the few data points that you have. (The % increase should be negative, as the temperature is decreasing.) The data only has 2 significant figures and you only have 5 data points. So I would say that the percent change in T is essentially constant when going from one time to the next.

I don't know at what level you are doing data analysis in your class. What are the ways that you have been taught to tell if a data set represents exponential behavior?
 
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  • #29
TSny said:
Well, as you stated in your first post, the percent increase is fairly consistent at -14% to -15% for the few data points that you have. (The % increase should be negative, as the temperature is decreasing.) The data only has 2 significant figures and you only have 5 data points. So I would say that the percent change in T is essentially constant when going from one time to the next.

I don't know at what level you are doing data analysis in your class. What are the ways that you have been taught to tell if a data set represents exponential behavior?
I based my answer on this video. In the example the change is constantly equal to 50 %. In my case the percentage differs (slightly, but differs), so I would assume that it is not exponential.

And in terms of (b) (ii)? Should the temperature of 29 degrees be correct?
 
  • #30
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.
 
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  • #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?
 
  • #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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Related to Calculating Energy and Mass in Espresso Machine and Analyzing Cooling of Coffee

1. What is the purpose of calculating energy and mass in an espresso machine?

The purpose of calculating energy and mass in an espresso machine is to understand the efficiency and performance of the machine. By knowing the amount of energy and mass involved in the process, we can determine the ideal settings for the machine and make adjustments to improve its performance.

2. How is energy and mass calculated in an espresso machine?

Energy and mass can be calculated using various equations and measurements. The energy input can be determined by measuring the power consumption of the machine and the time it takes to brew a shot of espresso. Mass can be calculated by weighing the amount of coffee beans used and the weight of the final product.

3. What factors affect the cooling of coffee in an espresso machine?

The cooling of coffee in an espresso machine is affected by several factors, including the temperature of the water used, the type and size of the coffee beans, the brewing method, and the temperature of the surrounding environment. Additionally, the design and efficiency of the machine's heating and cooling systems can also impact the cooling rate of the coffee.

4. How does the cooling of coffee affect its taste?

The cooling of coffee can significantly affect its taste. As coffee cools, its chemical composition changes, which can alter the flavor and aroma. Rapid cooling can lead to a more acidic and bitter taste, while slower cooling can result in a smoother and sweeter flavor. The rate of cooling can also impact the extraction of flavors from the coffee beans, ultimately affecting the overall taste of the coffee.

5. How can the cooling rate of coffee be improved in an espresso machine?

To improve the cooling rate of coffee in an espresso machine, several factors can be considered. These include using colder water, preheating the cup, using a larger surface area for the coffee, and adjusting the brewing method to allow for slower extraction. Additionally, regularly cleaning and maintaining the machine's heating and cooling systems can also improve its efficiency and, therefore, the cooling rate of the coffee.

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