Water has the following thermodynamic constants: (1) specific heat liquid = 4.18
ID: 994833 • Letter: W
Question
Water has the following thermodynamic constants: (1) specific heat liquid = 4.18 J/g °C, solid = 2.09 J/g °C, gas = 1.89 J/g °C, (2) heat of fusion = 334 J/g, and (3) heat of vaporization = 2257 J/g. For a sample of water at 1.0 atm of pressure, mass = 211 g at an initial temperature of -43 °C and a final temperature of 260 °C, answer the following questions:
(1) how much heat is required to warm the solid sample to its melting point? J
(2) how much heat is required to melt the sample? J
(3) how much heat is required to warm the liquid sample to its boiling point? J
(4) how much heat is required to vaporize the sample? J
(5) how much heat is required to warm the gaseous sample to its final temperature? J
(6) how much heat is required for the entire process to occur? J
Explanation / Answer
1)
q = mice*sice*DT
= 211*2.09*(0--43)
= 18962.57 joule
= 18.962 kj
2) q melting = DHfus*mass of sample
= 334 * 211 = 70474 joule
3) q = mwater*swater*DT
= 211*4.18*(100-0)
= 88198 joule
4) q vaporizing = DHvap*mass of sample
= 2257*211 = 476227 joule
5) q = mvap*Svap*DT
= 211*1.89*(260-100)
= 63806.4 joule
6) qabsorbed total = 18962+70474+88198+476227+63806.4 = 717667.4 joule
= 717.667 kj
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