Question about Entropy?

Question

Calculate the change in entropy of 230 g of water heated slowly from 25.0°C to 90.0°C.

I'm assuming I'm supposed to use dQ = mcdT but I'm not really sure which numbers to plug in where. The m is the mass of .23 kg, and I think c was 4.186 J/Kg*C. What would the dT be though?

(I tried just subtracting them out and got a change of 65*). Then plugged that in to get a dQ of 62.5807.

And then do I use S = dQ/T like:

dQ/T(final) - dQ/T(initial) = change in entropy?

That got me a final answer of .037569 which was incorrect even when cutting it to significant figures Any help would be appreciated =)

Answer 1

dT=90-25

dQ = mcdT
dQ= .23 x 4.186 x(90-25)=62.581 joules

Looks good so far

Now S=dQ(T2-T1)/(T2T1)
T - degrees Kelvin

S=62.581 (90-25)/[(273+90)(273+25)]
=0.0376joles/deg K

I'm with you