Specific Heat Capacity Question Mixture Q=MCdeltaT

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SUMMARY

The discussion revolves around calculating the specific heat capacity of an unknown liquid mixed with water. The mixture consists of 75g of the unknown liquid at 25 degrees Celsius and 60g of water at 90 degrees Celsius, resulting in a final temperature of 65 degrees Celsius. The specific heat capacity of the unknown liquid is calculated to be approximately 2.0905 J/g°C using the formula Q = mcΔT, where the heat lost by water equals the heat gained by the unknown liquid. The conclusion suggests that the unknown liquid could be steam or ice, but the temperature conditions indicate it must be a liquid, prompting further investigation into specific heat tables.

PREREQUISITES
  • Understanding of specific heat capacity and its calculation using Q = mcΔT
  • Basic knowledge of thermodynamics and heat transfer principles
  • Familiarity with temperature scales and phase changes of water
  • Access to specific heat capacity tables for various substances
NEXT STEPS
  • Research the specific heat capacities of common liquids to identify potential candidates for the unknown liquid
  • Learn about the effects of temperature on specific heat capacity and when it is appropriate to assume constancy
  • Explore phase change phenomena and their impact on specific heat calculations
  • Investigate the critical point of substances and how it affects heat capacity
USEFUL FOR

Students studying thermodynamics, chemistry enthusiasts, and educators looking to enhance their understanding of heat transfer and specific heat capacity calculations.

dav1d
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Homework Statement



A mixture is made by adding 75g of an unknown liquid at a temperature of 25 degrees Celsius to 60g of water at a temperature of 90 degrees Celsius. The final temperature f the mixture is 65 degrees Celsius. Calculate the specific heat capacity of the liquid. What is the liquid? How are you sure that you are correct?

Homework Equations


Q=m*c*deltaT


The Attempt at a Solution


I don't even know where to start. You can't calculate Q for water because C is not 4.181 at 90 degrees C. You can't find the Q of the unknown liquid because we do not know it's c value...
 
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You are correct that the heat capacity of a substance (C) varies with temperature. However C does not change very much with temperature, so unless we are doing very precise calculations, we usually just assume stays constant at all temperatures.
 
Ok, so in that case/

Is it correct to find Q for water with c=4.181, delta T = -25 degrees C

set that equal to mcdeltaT for the unknown liquid, rearrange and solve for c of the unknown and then match it up with a table??
 
That sounds like a good plan.
 
So c, specific heat capacity, changes only slightly between temperatures, so we can assume they are the same always? When does this not work?
 
Unless the material changes phase (i.e. liquid water turning into ice or vapor), it's a pretty good assumption that the heat capacity will not change significantly.
 
See for example this table. Specific heat of liquid water between 0°C and 100°C doesn't vary by more than 1%.

Apparently even without phase change differences can be substantial when temperature approaches critical point.
 
Ok so
Qwater=mcdeltaT
=60g*4.181J/gdegree C*(65-90degree C)
=-6271.5J

Qunknown=6271.5J (flip signs since the energy is absorbed by the unknown liquid)
c=-6271.5J/(75g)(65-25degree C)
=2.0905J/gdegree C

Now I don't know what it is, according to wikipedia tables for specific heat capacity it should be steam? or is it ice? But the question says liquid...
 
Last edited:
Do you have a table with specific heats in your textbook, notes, course materials? If so, look there.
 
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Borek said:
Do you have a table with specific heats in your textbook, notes, course materials? If so, look there.

I have, but it's pretty much split between ice or steam, which makes no sense since the temperature is too low to form steam and too high to form ice.
 

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