# Change in Entropy of water

1. Jan 19, 2005

### Kawrae

>> 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?

2. Jan 19, 2005

### Kawrae

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??

3. Jan 19, 2005

### learningphysics

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?