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The heat of combustion of ethane gas at 373 k cal per mole. Assume that 50% of heat is useful, how many be burnt litres to convert 60 kg of water at `20^(@)C` to steam at `100^(@)C` ? One mole of gas oc cupies 22.4 litre at S.T.P. Latent heat of steam = `2.25 xx 10^(6) Jkg^(-1)`. |
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Answer» Heat required to convert 60 kg of water at `20^(@)C` into steam at `100^(@)C` `=ms_(w) DeltaT + mL` `= 60 xx 10^(3) xx (100 -20) + 60 xx (2.25 xx 10^(6))/(4.2) cal` `= 4.8 xx 10^(6) + 32.14 xx 10^(6) = 36.94 xx 10^(6) cal` Since `50%` of heat is useful, so that heat produced is `Q = 100/50 xx 36.94 xx 10^(6) = 73.88 xx 10^(6) cal` Heat of combustion of ethane gas `=373 kcal mol^(-1) = 373 xx 10^(3) cal mol^(-1)` No. of moles of ethane required to be burnt for heat `Q is , n = (73.88 xx 10^(6))/(373 xx 10^(3)) = 198 mol e` Volume of ethane needed to be burnt ` 22.4 xx n` ` = 22.4 xx 198` ` =4435.2 litres`. |
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