Asked by Jeff
How much energy is required to bring 10.0 grams of ice at -5.00 degrees to 105-degree steam? Assume the system is isolated so that no mass is lost to the surroundings
Answers
Answered by
Damon
nobody knows because you did not give me:
specific heat of ice
heat of fusion of water
specific heat of water
heat of vaporization of water
Cp of steam
Energy =
10 grams (or .010 kg if using kg)
times the sum of:
[5 deg * spec heat ice
+ heat of fusion of water
+ 100 deg * spec heat of water
+ heat of vaporaization of water
+ 5 deg * specific heat of steam at one atmosphere ]
specific heat of ice
heat of fusion of water
specific heat of water
heat of vaporization of water
Cp of steam
Energy =
10 grams (or .010 kg if using kg)
times the sum of:
[5 deg * spec heat ice
+ heat of fusion of water
+ 100 deg * spec heat of water
+ heat of vaporaization of water
+ 5 deg * specific heat of steam at one atmosphere ]
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.