Using the vlues for the heat of fusion, specific heat of water, and/or heat of v
ID: 104477 • Letter: U
Question
Using the vlues for the heat of fusion, specific heat of water, and/or heat of vaporization, calculate the amount of heat energy in each of the following:
A) Kilocalories needed to melt a 525g ice sculpture at 0 degrees C and to warm the liquid to 15.o degrees C.
B) Kilojoules released when 85.0g of steam condenses at 100 degrees C, cools ad freezes at 0 degrees C.
Please show the formula and how so I can understand. For some reason I am not coming out with the correct answer and I don't know where I am messing up at.
Explanation / Answer
Q = heat change for conversion of ice at 0oC to water at 0oC + heat change for conversion of water at 0oC to water at 15oC
Amount of heat absorbed , Q = mL + mc'dt
= m(L + c'dt' )
Where
m = mass of ice = 525g
c” = Specific heat of steam = 2.1 J/ goC
c' = Specific heat of water = 4.186 J/g degree C
c = Specific heat of ice= 2.09 J/g degree C
L’ = Heat of Vaporization of water = 2260 J/g
L= Heat of fusion of ice = 334.9 J/g
dt' = 15-0=15 oC
Plug the values we get Q = m(L + c'dt ')= 209*10^3 J
Q= 209 kJ*(0.239kCal/1kJ)
Q=50.0 kCal
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Q = heat change for conversion of steam at 100 oC to water at 100 oC + heat change for conversion of water at 100oC to water at 0 oC + heat change for conversion of water at 0oC to ice at 0oC
Amount of heat released , Q = mL' + mc'dt' + mL
= m(L' + c'dt' + L )
m=. 85.0 g ; dt'= 100-0=100 oC ;
Q = 256*10^3 J= 256 kJ
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