1.Calculate the amount of energy (in kJ) necessary to convert 377 g of liquid wa

Question

1.Calculate the amount of energy (in kJ) necessary to convert 377 g of liquid water from 0 degree C to water vapor at 167 degress C. The molar heat of vaporization (Hvap) of water is 40.79 kJ/mol. The specific heat for water is 4.184 J/g degrees C, and for steam is 1.99 J/g degrees C. (Assume that the specific heat values do not change over the range of temperatures in the problem.)

2.Which substance has the highest vapor pressure at room temperature?

HF

HCl

HBr

HI

All of these substances have the same vapor pressure at room temperature.

Explanation / Answer

Q1

We require 2 type of heat, latent heat and sensible heat

Sensible heat (CP): heat change due to Temperature difference

Latent heat (LH): Heat involved in changing phases (no change of T)

Then

Q3 = m*Cp wáter * (Tb – Tf)

Q4 = m*LH vap

Q5 = m*Cp vap* (T2 – Tb)

Note that Tf = 0°C; Tb = 100°C, LH ice = 334 kJ/kg; LH water = 2264.76 kJ/kg.

Cp ice = 2.01 J/g°C ; Cp water = 4.184 J/g°C; Cp vapor = 2.030 kJ/kg°C

Then

Q3 = m*4.184 * (100 – 0)

Q4 = m*2264.76

Q5 = m*2.03* (T2 – 100)

QT = 377*(4.184 * (100 – 0) + 2264.76+2.03* (167– 100))

QT = 1062827.09 J = 1062.83 kJ

Q2

highest vapor pressure --> the heaviest hydrogen halide

this is HI, since it is present as a solid, which implies lowers vapor pressure

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