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

AI Thread Summary
To calculate the change in entropy of 210 g of water heated from 25.0°C to 70.0°C, the correct approach involves recognizing that the temperature changes during the heating process. The initial calculation used dQ = mcdT, yielding a value of 39.5570 J, but applying dS = dQ/T assumes a constant temperature, which is incorrect. Instead, the integration of dS = mcdT/T is necessary to account for the varying temperature. This requires integrating over the temperature range from the initial to final states. The discussion highlights the importance of using the correct method for calculating entropy changes in systems with temperature variations.
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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