What is the Correct Equation for Calculating Change in Entropy of Water?

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SUMMARY

The correct method for calculating the change in entropy of water involves integrating the equation dS = dQ/T, considering the variable temperature during the heating process. For 210 g of water heated from 25.0°C to 70.0°C, the specific heat capacity (c) is 4.186 J/g·K. The calculation requires the integration of dS = mc * ln(Tf/Ti) rather than assuming a constant temperature, which was the initial error in the discussion. The final result for the change in entropy is derived from the integration of the heat transfer over the temperature range.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically entropy.
  • Familiarity with the specific heat capacity of water (4.186 J/g·K).
  • Knowledge of calculus, particularly integration techniques.
  • Ability to convert temperatures from Celsius to Kelvin.
NEXT STEPS
  • Study the integration of thermodynamic equations, focusing on variable temperature scenarios.
  • Learn about the derivation of the entropy formula dS = mc * ln(Tf/Ti).
  • Explore practical applications of entropy calculations in real-world thermodynamic systems.
  • Review examples of calculating changes in entropy for different substances and conditions.
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Students and professionals in physics and chemistry, particularly those studying thermodynamics and heat transfer principles.

Kawrae
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>> Calculate the change in entropy of 210 g of water heated slowly from 25.0°C to 70.0°C. (Hint: Note that dQ = mcdT.) <<

I'm not sure how to start this, but I think I need to use the equation:
dS=m1*c1*ln(Tf/T1) + m2*c2*ln(Tf/T2)

I know c of water = 4.186J and I know m1=m2... but I don't have a second temperature to use to solve the equation. But this is the only change in entropy equation I know of... am I missing something? :frown:
 
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Okay, I found part of my mistake but now I don't know why this isn't getting the correct answer...

I used dQ = mcdT and plugged in:
dQ = (.210kg)*(4.186J)*(45K)
dQ = 39.5570J

Then I used...
dS = dQ/T
Plugging in...
dS = 39.5570/343.15K
dS = .115J/K

What am I doing wrong??
 
Kawrae said:
Okay, I found part of my mistake but now I don't know why this isn't getting the correct answer...

I used dQ = mcdT and plugged in:
dQ = (.210kg)*(4.186J)*(45K)
dQ = 39.5570J

Then I used...
dS = dQ/T
Plugging in...
dS = 39.5570/343.15K
dS = .115J/K

What am I doing wrong??

Remember that the formula dS=dQ/T is at a specific temperature for an infintesimal change in heat. What you did above is incorrect as it assumes that temperature is constant.

dQ=mcdT

dS=dQ/T

so dS=mcdT/T

Integrate both sides... what do you get?
 

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