Determining Liters of ethane burned to convert H20 to steam

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The discussion focuses on calculating the liters of ethane needed to convert 50 kg of water from 10°C to steam at 100°C, using the heat of combustion of ethane. The heat required is calculated as 31,500 calories, leading to a mole calculation based on the heat of combustion, which is 373 kcal/mole with 60% efficiency. The initial calculation yielded 3,193,680 liters, which was questioned against a textbook answer of 315,000 liters. Participants identified potential errors in unit conversions and the application of the ideal gas law, particularly regarding the use of moles and the volume occupied by gas at standard conditions. The discussion highlights the importance of careful unit management in thermodynamic calculations.
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Homework Statement


The Heat of combustion of Ethane is 373 kcal/mole. Assuming that 60% of the heat is useful , how many liters of ethane,, measured at standard temperature and pressure, must be burned to convert 50.0kg of water ar 10.0C to steam at 100.0C? One mole of a gas occupies 22.4 liters at precisely 0C and 1atm.

Homework Equations


PV=nRT Q=mcΔT

The Attempt at a Solution


Qburned=Q heat up

=mL + mcΔT

540g/C(50g)+ 50(1cal/gC)(100-10)
Q=31500 cal

n= Q/heat of combustion(60%)
n=0.1407mol

PV=nRT
V=0.1407mol(8314j/kmolK)(273K)/1atm

V=3193680 L

The book has it at 315000. Is there something wrong here? I feel like I am missing a term or left something out?
 
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You calculated Qburned per 50 grams. Did you convert this to per 50 kg somewhere?
 
Why use PV=nRT when you're given 22.4 L/mol?
 
Ahh that's where the error comes from. I forgot that it is per mole
 
NascentOxygen said:
You calculated Qburned per 50 grams. Did you convert this to per 50 kg somewhere?
I did I believe. It should be in my note pad
 

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