The heat of fusion for water is the energy change at 0 degree C for the followin
ID: 1065804 • Letter: T
Question
The heat of fusion for water is the energy change at 0 degree C for the following: H_2O(s) rightarrow H_2O(l). For water, the heat of fusion is 6.02 kJ/mol. If 5.00 g of ice at 0.00 degree C is added to 100.0 g of water, initially at 20.00 degree C in a calorimeter, how much will the temperature of the initially 20.00 degree C water change? The specific heat of liquid water is 4.18 J/g degree C, and the heat capacity of the calorimeter is 21.0 J/degree C. To 50.0 g of water at 25.00 degree C in a calorimeter was added 5.00 g of solid calcium chloride, also at 25.00 degree C. The specific heat of the resultant solution was 4.20 J/g degree C. The heat capacity of the calorimeter was 20.0 J/degree C. The energy change for dissolving solid calcium chloride in water is -81.5 kJ per mole of calcium chloride. Write the chemical equation showing the dissolution of solid calcium chloride in water. Is the dissolution of solid calcium chloride in water endothermic, exothermic, neither, or cannot be determined? Justify your choice. After the solid calcium chloride dissolved in water, what was the final solution temperature? Two samples of water are heated from 20.0 degree C to 60.0 degree C. One of the samples required twice as much heat to bring this temperature change as the other. How do the masses of the two water samples compare? Explain your reasoning.Explanation / Answer
Answer (2)
First if all , it is very important that you do not forget to account for the phase change underwent by the solid water at 0C to liquid at 0C.
The heat needed to melt the solid at its melting point will come from the warmer water sample. This means that you have
q1+q2=q3 (1), where
q1 - the heat absorbed by the solid at 0C
q2 - the heat absorbed by the liquid at 0C
q3 - the heat lost by the warmer water sample
The two equations that you will use are
q=mcT , where
q - heat absorbed/lost
m - the mass of the sample
c - the specific heat of water, equal to 4.18 J/gC
T - the change in temperature, defined as final temperature minus initial teemperature
and
q = nHfus , where
q - heat absorbed
n - the number of moles of water
Hfus - the molar heat of fusion of water, equal to 6.02 kJ/mol
Use water's molar mass to find how many moles of water you have in the 5.0 gm sample
5.0 gm x ( 1 mole H2O / 18.015 gm ) = 0.2775 moles H2O
So, how much heat is needed to allow the sample to go from solid at 0C to liquid at 0C?
q1 = 0.2775 moles x 6.02 kJmole = 1.6708 kJ
This means that equation (1) becomes
1.6708 kJ + q2 = q3
The minus sign for q3 is used because heat lost carries a negative sign.
So, if Tf is the final temperature of the water, you can say that
1.6708 kJ + Msample x c x Tsample=Mwater x c x Twater
1.6708 kJ+ ([5 x 4.18 x (Tf0) ]= - [100 x 4.18 x (Tf20) ]
1.6708 kJ+ ([20.9 J x (Tf) ]= - 418 J x (Tf20)
1.6708 kJ+ 0.0209 KJ (Tf) = - 0.418 KJ Tf + 8.360
0.4389 Tf = 6.689
Tf = 15.24 C
Rounded to two sig figs, the number of sig figs you have for the mass of warmer water, the answer will be
Tf = 15C
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.