Hello, this is from a heat of fusion and vaporization lab that I am having a bit

Question

Hello, this is from a heat of fusion and vaporization lab that I am having a bit of difficulty with. I posted this question twice, but the 1st response I received was illegible,and the 2nd did not know how to answer the question. Then no one else answered afterwards. ALL OF THE DATA IS PROVIDED. I've answered all questions on the lab except for this one.

If you were to double the amount of ice and also double the amount of hot water in the first experiment to determine the heat of fusion, how would it change the final tempurature of the mixture? (Assume the stirrer takes up an insignificant amount of energy.) Reason your answer without doing any calculations.

My values for the first experiment:

Mass of calorimeter and water: 0.258kg

Mass of water: 0.239kg

Mass of water, ice, calorimeter: 0.275kg

Mass of ice: 0.036kg

Mass of calorimeter: 0.0118kg

Mass or calorimeter stirrer: 0.0278kg

specific heat of stirrer cstr: 910 J/kg K

Temp of hot water TH: 37.8 oC

Temp of ice T0: 0 oC

Final temp T: 33.3 oC

Latent heat of fusion (from my own calculation...): 7666.5 J/kg

Explanation / Answer

I am not familiar with the first experiment (that you mention) of how did you determine the heat of fusion but i have the feeling that the final temperature of the mixture will not change because is function of the mass proportion. You might get a different value for the heaf of fusion but if both components of the mixture are doubled i believe you will have the same final temperature

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.