1) The following information is given for n -pentane at 1 atm: A 40.00 g sample
ID: 707267 • Letter: 1
Question
1) The following information is given for n-pentane at 1 atm:
A 40.00 g sample of liquid n-pentane is initially at –27.30°C. If the sample is heated at constant pressure ( = 1 atm), _______kJ of energy are needed to raise the temperature of the sample to 53.90°C.
2) The following information is given for magnesium at 1 atm:
A 48.10 g sample of solid magnesium is initially at 637.00°C. If the sample is heated at constant pressure ( = 1 atm), ________kJ of heat are needed to raise the temperature of the sample to 702.00°C.
3)The following information is given for tin at 1 atm:
A 28.40 g sample of solid tin is initially at 207.00°C. If J of heat are added to the sample at constant pressure ( = 1 atm), which of the following is/are true?
(Select all that apply.)
The sample is a liquid.
The sample is a solid in equilibrium with liquid.
The sample is at a temperature greater than 232.00°C.
The sample is at a temperature of 232.00°C.
The sample is a solid.
= 36.20°C (36.20°C) = J/g = –129.70°C (–129.70°C) = J/g Specific heat gas = 1.650 J/g °C Specific heat liquid = 2.280 J/g °CExplanation / Answer
Ans 1
Data for n-pentane at 1 atm
boiling point Tb = 36.20 °C
Latent heat of vaporization
Hvap(36.20 °C) = 357.6 J/g
melting point Tm = -129.70 °C
Latent heat of fusion
Hfus(-129.70 °C) = 116.7 J/g
specific heat gas Cpg = 1.650 J/g°C
specific heat liquid Cpl = 2.280 J/g°C
Mass of liquid n-pentane, m = 40 g
Initial temperature T1 = - 27.30°C
n-pentane is in liquid state initially
Heat required to change the temperature from - 27.3 to 36.2c
Q1 = m x Cpl x (T2-T1)
= 40 g x 2.280 J/gK x (36.2 + 27.3)K
= 5791.2 J
Heat required to change the phase from liquid to gas
Q2 = m x Hvap
= 40 g x 357.6 J/g = 14304 J
Heat required to change the temperature from 36.2 to 53.9c
Q3 = m x Cpg x (T2-T1)
= 40 g x 1.650 J/gK x (53.9 - 36.2) K
= 1168.2 J
Total heat required
= 5791.2 + 14304 + 1168.2
= 21263.4 J
= 21.263 kJ
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