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

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SUMMARY

The discussion focuses on calculating the change in entropy when 27 grams of ice at 0°C is added to 250 grams of water at 23°C. The final temperature is determined to be 20.75°C using the heat transfer equation, where the heat lost by the water equals the heat gained by the ice. The change in entropy is calculated using the formula ΔS = ∫ dQrev/T, leading to a result of 0.78 calories/°C. The key steps involve calculating the heat required to melt the ice before raising its temperature.

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  • Understanding of thermodynamics concepts, specifically entropy.
  • Familiarity with the heat transfer equation and specific heat capacity.
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  • Study the principles of thermodynamics, focusing on entropy calculations.
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  • Explore the concept of reversible processes in thermodynamics.
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devil0150
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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?
 
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devil0150 said:
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 = dQrev/T or ΔS = ∫ dQrev/T

where dQrev 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
 

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