ProdigyInspired
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- 2016
A student used the apparatus to determine the molar heat of combustion of propanol.
Results:
Mass of 1-propanol burnt = 0.60g
Mass of Water heated = 200g
Initial Temp of Water = 21 C
The molar heat of combustion of 1-propanoil is 2021 kJ mol-1, Assuming no heat loss, what would be the final temperature of the water?
My Attempt:
AFIK we have to find T from the q=mCT equation.
\quad=\quad \frac { m }{ mm } \quad =\quad \frac { 0.60g }{ 60 }=\quad 0.01\quad \\ using\quad \triangle H\quad =\quad \frac { q }{ n } \\ q\quad =\quad 0.01\quad \times \quad 2021\quad kJ\quad =\quad 20.21\quad kJ\quad =\quad 20210\quad J\\ using\quad q\quad =\quad mC\triangle T\\ \triangle T\quad =\quad \frac { 20210\quad J }{ 4.18\quad \times \quad { 10 }^{ 3 }\quad \times \quad 200g } =\quad 0.02417..)
The answer for some reason omits the 10^3 from the denominator, giving it to be around 24 C, therefore the final is C, 45 degrees.
EDIT: I found my 'error'. The mass described in q=mCT seems to be separate, so it's in kg rather than g corresponding to the molar mass found. So when you use the n = m/mm formula and use grams, you don't convert it into kg?
Results:
Mass of 1-propanol burnt = 0.60g
Mass of Water heated = 200g
Initial Temp of Water = 21 C
The molar heat of combustion of 1-propanoil is 2021 kJ mol-1, Assuming no heat loss, what would be the final temperature of the water?
My Attempt:
AFIK we have to find T from the q=mCT equation.
The answer for some reason omits the 10^3 from the denominator, giving it to be around 24 C, therefore the final is C, 45 degrees.
EDIT: I found my 'error'. The mass described in q=mCT seems to be separate, so it's in kg rather than g corresponding to the molar mass found. So when you use the n = m/mm formula and use grams, you don't convert it into kg?
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