Answer: 3.66 x 10^6 J or 874.760994 kilocalories. Question : one kilogram of ice

Question

Answer: 3.66 x 10^6 J or 874.760994 kilocalories. Question : one kilogram of ice at -20 degree Celsius is heated to 0 degrees Celsius and melted. The resulting water is heated to 100degrees celcius, vaporizes, and the steam is further heated to 400 degrees Celsius. Assume that all the processes are isobaric. Calculate the change in enthalpy in kilocalories and in joules. (The specific heats at constant pressure of ice, liquid water, and steam are 0.55, 1.00, and 0.48 kcal kg^-1, respectively.) I need for the question to be solved where my answer is the one above!

Explanation / Answer

Q1 = 1*0.55*20=11 kcal

latent heat of ice =79.72 kcal/kg

Q2 = 1*79.72 = 79.72 kcal

Q3 = 1*1.00*100 = 100 kcal

latent heat of vaporisation of water = 539 kcal/kg

Q4 = 1*539 = 539 kcal

Q5 = 1*300*0.48 = 144 kcal

Q1+Q2+Q3+Q4+Q5 = 11+79.72+100+539+144 = 873.72 kcal

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