- #1
devil0150
- 3
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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?
(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?