Specific Heat Capacity Question Mixture Q=MCdeltaT

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Discussion Overview

The discussion revolves around a homework problem involving the calculation of the specific heat capacity of an unknown liquid mixed with water. Participants explore the implications of temperature on specific heat capacity and the methodology for solving the problem, including the application of the heat transfer equation Q=mcΔT.

Discussion Character

  • Homework-related
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant expresses uncertainty about starting the calculation due to the variable nature of specific heat capacity with temperature.
  • Another participant suggests that specific heat capacity can be assumed constant for practical calculations unless high precision is required.
  • A participant proposes calculating the heat transfer for water using a specific heat value of 4.181 J/g°C and setting it equal to the heat transfer for the unknown liquid to solve for its specific heat capacity.
  • There is a discussion about the conditions under which the assumption of constant specific heat capacity holds, particularly noting that phase changes significantly affect this assumption.
  • One participant calculates the specific heat capacity of the unknown liquid to be approximately 2.0905 J/g°C but is unsure about the identity of the liquid based on standard tables.
  • Another participant suggests checking course materials for specific heat values, indicating that the results may not align with common expectations.

Areas of Agreement / Disagreement

Participants generally agree that specific heat capacity can be approximated as constant for small temperature ranges, but there is no consensus on the identity of the unknown liquid or the appropriateness of the calculated value based on the given temperatures.

Contextual Notes

Participants note that the specific heat capacity of water does not vary significantly between 0°C and 100°C, but there are concerns about the accuracy of this assumption near critical points or during phase changes.

Who May Find This Useful

This discussion may be useful for students studying thermodynamics, specifically those dealing with heat transfer calculations and the properties of materials in mixtures.

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.
 
  • #10
Ʃ
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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