How Do You Calculate the Change in Entropy When Ice Melts in Water?

[devil0150]

In 250gr water at 23 oC(degrees celsius) we throw in 27gr ice at 0 oC. Find the change in entropy.
(sorry if that was a bad translation but English is not my native language)
The answer is 0.78 calories/oC. But I'm not sure how do this. The formula I have for entropy is: S=δQ/T

I find the final temperature like this: (q lost in water = q gained in ice)
m1 * 4.184 * (23 - ft) = m2 * 4.184 * (ft - 0)
ft = 20.75 oC

Now how do I find δQ?

 

[Andrew Mason]

In 250gr water at 23 oC(degrees celsius) we throw in 27gr ice at 0 oC. Find the change in entropy.
(sorry if that was a bad translation but English is not my native language)
The answer is 0.78 calories/oC. But I'm not sure how do this. The formula I have for entropy is: S=δQ/T

The formula for change in entropy between two states is: dS = dQ_rev/T or ΔS = ∫ dQ_rev/T

where dQ_rev is the heat flow over a reversible path between those states.

I find the final temperature like this: (q lost in water = q gained in ice)
m1 * 4.184 * (23 - ft) = m2 * 4.184 * (ft - 0)
ft = 20.75 oC

Now how do I find δQ?

Before you can raise the temperature of the ice, you first have to melt it. What is the heat required to melt 27 g of ice?

AM