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?