Calcuate the enthalpy change delta H, for the process in which 37.7 g of water i
ID: 978780 • Letter: C
Question
Calcuate the enthalpy change delta H, for the process in which 37.7 g of water is coneverted from liquid at19.9 degree C to vaper at 25.0 degree C. For water, delta H_vap = 44.0 KJ/mol at 25.0 degree C and C_n = 4.18 J/(g middot degree C) for H_2O(l). Express your answer lo three significant figures and include the appropriate units. submit Part B How many grams of ice at -10.3 degree C can be completely coneverted to liquid at 21.0 degree C. If the available heat for this process is 5.53 times 10^3 kJ? For ice, use a specific heat of 2.01 J/(g middot degree C) and delta H_m = 6.01 kJ/mol Express your answer lo three significant figures and include the appropriate units. submitExplanation / Answer
In the part A it is mentioned that the conversion of vapour at 25 oC , it is impossible.So correct the question please.
Part B:
Q = heat change for conversion of ice at -10.3 oC to ice at 0 oC + heat change for
conversion of ice at 0oC to water at 0oC + heat change for conversion of water at 0oC to water at 21.0 oC
Amount of heat absorbed , Q = mcdt + mL + mc'dt
= m(cdt + L + c'dt' )
Where
m = mass of ice = ?
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 fusion of ice = 334.9 J/g
dt' = 21.0 -0 = 21.0oC
dt = 0-(-10.3) = 10.3 oC
Q = amount of heat available = 5.53x103 kJ
= 5.53x106 J
Plug the values we get Q = m(cdt + L + c'dt')
m = Q / (cdt + L + c'dt')
= 12.45x103 g
Therefore the mass of ice required is 12.45x103 g