Determining Liters of ethane burned to convert H20 to steam

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Homework Help Overview

The problem involves calculating the amount of ethane needed to convert a specific mass of water from a lower temperature to steam, focusing on the heat of combustion and the efficiency of heat use. The subject area includes thermodynamics and combustion chemistry.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of heat required to convert water to steam and the conversion of units from grams to kilograms. Questions arise regarding the use of the ideal gas law versus molar volume at standard conditions.

Discussion Status

The discussion is ongoing, with participants identifying potential errors in calculations and questioning the assumptions made in the original poster's approach. There is no explicit consensus yet, but some guidance has been offered regarding unit conversions and the application of formulas.

Contextual Notes

Participants note the importance of efficiency in heat transfer and the specific conditions under which the calculations are made, including standard temperature and pressure for gas measurements.

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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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