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 was able to answer all but this question.

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

Without calculations, heat losty by water and acquired by ice melting in water depends linearly on mass, so if I double both mass and thus heats and them equal them to compute equilibrium tempreature I guess nothing would change with the case with half masses.

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