Discover the best answers to your questions with the help of IDNLearn.com. Get prompt and accurate answers to your questions from our community of knowledgeable experts.
The final stabilized temperature will be between 0 °C and 50 °C.
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
#SPJ4