1. Use given values as stated in questions below. 2. Clearly State Answer 3. (Pl
ID: 795736 • Letter: 1
Question
1. Use given values as stated in questions below.
2. Clearly State Answer
3. (Please) LEAVE A COMMENT if you are currently working on the problems.
4. See the following link for the thermo data if needed:
www.mediafire.com/download/t5ja3jz7a3vwb4t/Thermo_Data.docx
5. Points will be awarded if both anwers are correct.
4. Calculate the mass of methane that must be burned to provide enough heat to convert 330.0 g of water at 37.0
Explanation / Answer
4) Given 330 g of water at 37 C is converted to steam at 124 C
Total Heat required = heat required to convert water at 37 C to 100 C + heat required to convert water at 100 C to steam at 100 C + heat required to convert steam at 100 C to 124 C
total heat required = M Sw dT1 + ML + M Ss dT2
= 330 x 4.18 x (100-37) + 0.33 x 2260 x 103 + 330 x 1.865 x (124-100) )
=330 x 4.18 x 63 + 330 x 2270 + 330 x 2.08 x 24
= 850765.08 J
= 850.765 kJ
So total heat required is 850.765 kJ
The combustion reaction is
CH4 + 202 -----> C02 + 2H20
dH = dHf C02 + 2dHf0(H20) - dHf0 ( CH4)
= -393.5 -2 x 241.8 +74.81
= -802.29 kJ
1 mole of methane gives 802.29 kJ of heat
let y mole of methane give 850.765 kJ of heat
y= 850.765 /802.29
y= 1.00604
so 1.0604 moles of methane is required.
mass of methane = moles x M.W
=1.0604 x 16
= 16.97 g
So mass of methane required is 16.97 grams
17 ) We know that at freezing point
dHo = T dSo
T= dHo/dSo
where T is the freezing point in Kelvin
Given dHo= 0.260 kJ/mol
dSo= 4.851 J /mol K
so T= 0.26 x 1000/ 4.851
T = 53.60 K
Freezing point of fluorine is 53.60 Kelvin
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