Using the values for the heat of fusion, specific heat of water, and or heat vap
ID: 722213 • Letter: U
Question
Using the values for the heat of fusion, specific heat of water, and or heat vaporization, calculate the amount of heat energy in each of the following: a) joules to condense 125 g of steam at 100 degrees celsius and to cool the liquid to 15.0 degrees celsius; b) joules needed to melt a 525-g ice sculpture at 0 degrees celsius and to warm the liquid to 15.0 degrees celsius; c) kilojoules released when 85.0 g of steam condenses at 100 degrees C, cool the liquid, and freeze it at 0 degrees C; d) joules to warm 55.0 mL of water (density=1.00 g/mL) from 10.0 degrees C and vaporize it at 100 degrees C.Explanation / Answer
Data:
heat of vaporization = 44 KJ/mol
heat of fusion = 333.55 KJ/Kg
sp. heat of water = 4.186 KJ/Kg.K
(a) you have 125 gm steam at 100 degC to be cooled to liquid water at 15 degC
125gm of steam = (125/18) = 6.944 moles of steam
phase change involved = steam at 100 deg C to liquid water at 100 degC
liquid water at 100 degC to liquid water at 15 degC
so total heat change will be sum of heat changes of these two transitions.
heat "is evolved" during phase change involved of steam at 100 deg C to liquid water at 100 degC = 6.944*(heat of vaporization) = 305.56 KJ
heat "is evolved" when liquid water cools from 100 to 15 degC = m*(sp.heat)*dT
m = 125gm=0.125Kg, sp.heat = 4.186 KJ/Kg.K, dT = 100-15 = 85K
heat lost = 44.48KJ
net heat "lost" = 305.56 + 44.48 = 350KJ (since heat is lost, you may put a negative sign if the same is followed by you as convention)
(b) 525gm ice at 0 degC to liquid water at 15 degC
again two changes are there: ice at 0 degC to liquid water at 0 degC
and, water at 0 degC to water at 15 degC
mass of ice/water = 525gm = 0.525kg
heat absorbed while first chage (ice->water) = (heat of fusion)*(mass of ice) = 0.525*333.55 = 175.11KJ
heat absorbed for second change = m*c*dT
we have m=0.525kg, c = 4.186 KJ/Kg.K, dT = 15K... put all values to get heat absorbed = 32.96KJ
net heat "absorbed" = 175.11+32.96 = 208.07KJ (since heat is gained, you may put a positive sign if the same is followed by you as convention)
(the procedure is same for all cases... for parts (c) and (d) I am writing the changes involved, heat released/absorbed can be calculated as demonstrated.
(c)
85 gm of steam at 100 degC to liquid water at 100 degC
heat is "lost" and is equal to (85/18)*(44) = 207.78KJ
next
85gm or 0.085kg of liquid water at 100 degC is cooled to 0 degC
heat lost = m*c*dT = 0.085*4.186*100 = 35.6KJ
next
85gm of liquid water has to be frozen to ice at 0 DegC
heat lost = 0.085*333.55 = 28.35KJ
net heat lost = 271.82KJ (-ve sign may apply as heat is lost)
(d)
as density is 1gm/ml , 55ml weighs 55gm or 3.056 mol or 0.055kg
heat added when this is heated from 10 degC to 100 degC
heat added = m*c*dT = 0.055*4.186*90 = 20.72KJ
heat added when 55gm or 3.056mol of water is vaporized to steam at 100 degrees = (3.056*44) = 134.44KJ
net heat added = 20.72+134.44 = 155.16 KJ (+ve sign may apply as heat is added)
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