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If you mix 100 g of ice at 0 ∘c with 100 g of boiling water at 100 ∘c in a perfectly insulated container, the final stabilized temperature will be.

Sagot :

tushargupta0691

The final stabilized temperature will be between 0 °C and 50 °C.

Calorimetry:

The enthalpy of fusion of ice is 334 J/g. The specific heat of water is 4.2 J/g.

To cool 100 g of water from 100 °C to 0 °C would require the removal of

4.2 x 100 x 100 = 42000 J.

To melt the ice would require the addition of

334 x 100 = 33400 J

∴ 42000  > 33400 thus you can melt all the ice and have some heat to spare, specifically 42000 - 33400 = 8600 J

Now use this to warm up 100+100 = 200 g of water at 0 °C

The final stabilized temperature;

8600 / (200 x 4.2) = 10.23 °C

Therefore, the final stabilized temperature is 10.23 °C

Learn more about Calorimetry here:

https://brainly.com/question/1989313

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