- #1
Kawrae
- 46
- 0
>> 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?
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?