The following information is given for water at 1atm: boiling point = 100.0 degr
ID: 984376 • Letter: T
Question
The following information is given for water at 1atm: boiling point = 100.0 degree C Melting point = 0.000 degree C specific eat solid = 2.100 j/g degree C specific heat liquid = 4.184 j/g degree C delta H_vap(100.0 degree C) = 2.259 times 10^3 j/g delta H fos(0.000 degree C) = 333.5 j/g A 22.10 g sample of solid water is initially at -27.00 degree C. If 8.158 times 10^3 j of heat are added to the sample at constant pressure (P = 1 atm) which of the following are true. Choose all that apply: The samole is a gas The sample is a solid in equilibrium with liquid. The sample is at a temperature greater than 0.000 degree C. The sample is a liquid. The sample is at a temperature of 0.000 degree C.Explanation / Answer
the sample is solid at 0.000c
energy require to convert ice at -270 to ice at 00c = 22.10 x 2.1 x 27 =1253.7 J
energy required to convert ice at 00c to water at 00c = 22.1 x 333.5 =7370.35 J
energy we given only 8158 j. it is not enough to convert solid ice to liquid ice.
so sample solid at 00c
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