Question

The heat of combustion of ethane gas at 373 k cal per mole. Assume that 50% of heat is lost, how many litres of ethane measured at STP must be convert `50` kg of water at `10^(@)C` to steam at `100^(@)C` ? One mole of gas oc cupies `22.4` litres at STP. Take latent heat of steam `=2.25xx10^(6) j kg^(-1)` .

Answer 1

Correct Answer - [3758.2 litre]`
Total heat energy required to convert `50` kg of water at `10^(@)C` to steam at `100^(@)C` `=s m Delta T+ml`
`=1000xx50xx(100-10)+(50xx2.25xx10^(6))/(4.2)`
`4.5xx10^(6)+26.79xx10^(6)=31.29xx10^(6) cal.`
As, `50%` of heat is lost,
`:.` total heat produced `=2xx31.29xx10^(6) cal`.
Heat of combustion `=373xx10^(3) cal//mol e` .
`:.` Number of moles of ethane to be burnt
`=(2xx31.29xx10^(6))/(373xx10^(3))mol e`
Volume of ethane `=(2xx31.29xx10^(6)xx22.4)/(373xx10^(3))litre =3758.2 litre`