Calculating Mass Increase of Coffee Using Specific Heat Capacity

AI Thread Summary
The discussion revolves around calculating the mass increase of coffee when its temperature is raised from 34°C to 97°C using the specific heat capacity formula Q=mcΔt. Participants express confusion about the problem's clarity, particularly regarding the temperature of the coffee being added and how to determine the heat energy (Q). It is noted that if the added coffee is at the same temperature as the original, it won't increase the temperature. A participant successfully calculates the energy required to heat the coffee to 97°C as 8.85 x 10^4 J and suggests using E=mc² to find the equivalent mass increase. The conversation highlights the need for context in problem statements, especially in relation to special relativity.
Smartgurl
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Homework Statement



Using coffee (specific heat capacity of 3.98 x 10^3 J/kg °C) calculate the increase in mass if you raised the temperature of a 353 g cup of coffee from 34° C to 97°C.

Homework Equations


Q=mcΔt

The Attempt at a Solution


I believe I just need to solve for m. However I have no idea what my Q value is or how to find it. And for the change in temperature I'm not sure if I have to convert it to F or leave it in Celsius.
 
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Is this the entire question? You're increasing the temperature of 353 grams of coffee by pouring more coffee into the cup, right? Does the problem not tell you the temperature of the coffee you're adding? (Is it 100 degrees Celsius?)
 
Nathanael said:
Is this the entire question? You're increasing the temperature of 353 grams of coffee by pouring more coffee into the cup, right? Does the problem not tell you the temperature of the coffee you're adding? (Is it 100 degrees Celsius?)
Thats the whole question and all the info I know. It's part of an independent study so I have no examples or notes or anything to look off of. That's why I have no idea how to start it.
 
Well the problem doesn't make sense if you don't know the temperature of coffee that you're adding. The amount of coffee you need to add to the cup to raise it's temperature to 97°C depends on the temperature of the coffee you're adding.
For example, the temperature of the coffee already in the cup is 34°C, so if you're adding coffee that is also 34°C then you could add as much of it as you'd like, but it's not going to increase the temperature.

Let's just say the temperature of the coffee you're adding is 100°C, then do you have any ideas about how to approach the problem?
 
Smartgurl said:
It's part of an independent study
Could they be expecting you to calculate the mass equivalent of the added energy? It's an extremely flaky problem statement no matter how you look at it.
 
I am not sure the question is about adding coffee. It sounds like a typical question from an intro to special relativity. I would therefore like the OP to mention what kind of course this problem was given in for context.
 
The chapter is about special relativity. However I believed it had nothing to do with the unit because I don't see how it has anything to do with the topic. I don't remember reading anything about temperature in this topic and I don't think I have any equations that can work for this.
 
The problem is definitely coupled to the module. How much energy must you add to heat the water? What is the equivalent mass of this energy?
 
So I went back and read everything. Then I found the energy by using q=mcΔt and got 8.85 x 10^4 J. Now I think I use the equation E=mc^2 and solve for m.
 
  • #10
Smartgurl said:

Homework Statement



Using coffee (specific heat capacity of 3.98 x 10^3 J/kg °C) calculate the increase in mass if you raised the temperature of a 353 g cup of coffee from 34° C to 97°C.

Homework Equations


Q=mcΔt

The Attempt at a Solution


I believe I just need to solve for m. However I have no idea what my Q value is or how to find it. And for the change in temperature I'm not sure if I have to convert it to F or leave it in Celsius.

Look at the units for the given specific heat capacity of the coffee (3.98*103 J/kg-°C. Look at the units for temperature (34 °C to 97 °C). Why on Earth would you think a conversion of temperature from °C to °F would be required?
 

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