Question

2. How much energy (in kJ) is released when a 75.0 g of water vapor at 125°C forms an ice cube at -5.50°C? HINT: it still has to go through all of the same steps, even if they are not listed like in the last problem.

Answer 1

The water vapour has to undergo following changes

a) The water vapour comes to 100⁰C

b) The water vapour convertes to liquid at 100⁰C

c) As a liquid it goes to 0⁰C

d) At 0⁰C, it converts to ice (Solid)

e) As a solid it goes to 5.5 ⁰C

let us calculate the amount of heat released during this conversion using the following equation

Q = mc∆T

Q = heat energy (Joules, J), m = mass of a substance (g)

c = specific heat (units J/g∙^oC), ∆ is a symbol meaning "the change in"

∆T = change in temperature (^oC Celcius)

Heat of vaporization = 2259 J g¯¹

The heat of fusion for water at 0 °C is = 334 joules (79.7 calories) J g¯^1oC ,
specific heat capacity for solid water (ice) = 2.06 J g¯^1oC
specific heat capacity for liquid water = 4.184 J g¯^1oC
specific heat capacity for gaseous water (steam) = 2.02 J g¯^{1 o}C

Q = (75 g x 2.02 J g¯^{1 o}C x 25 ^oC ) + ( 75 g x 2259 J g¯¹ ) + ( 75 g x 4.184 J g¯^{1 o}C x 100 ^oC) + (75 g x 334 J g¯¹ ) + ( 75 g x 2.06 J g¯^{1 o}C x 5.5 ^oC)

Q =  3,787.5‬ J + 1,69,425‬ J + 31,380 J + 25,050‬ J + 849.75‬ J

Q =  2,30,492.25‬ Joules or  230.492 KJ

Hence 230.492 Kilo Joules of heat will be released. ‬