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Heya.......!!!!
first we need to take out No. of moles of Ethylene = 4.2/28 = 0.15 mole .
=> 1 mole on Combustion releases 340 kcal So ,, 0.15 mol will release = 340 × 1.5 = 51 kcal .
=> Energy required to melt 1000 g of ice at -10℃ ,,
By the general formulae = E = ms∆t = 1000 × 0.5 × 10 = 5 kcal .
Energy left for forming water = 51 - 5 = 46kcal.
Now ,, specefic latent heat of fusion of ice = 80cal/g , this 80 cal energy is required for fusion of 1 g ice .
=> 46000 cal energy will fuse x gram ice .
=> x = 46000÷80 = 575 g ice.
we know that the Density of water = 1 ( in CGS )
Hence ,, D = m/v = v = m/D = 575/1
➡ Volume of water available for drinking
= 575 ml
Hope It Helps You ^_^
first we need to take out No. of moles of Ethylene = 4.2/28 = 0.15 mole .
=> 1 mole on Combustion releases 340 kcal So ,, 0.15 mol will release = 340 × 1.5 = 51 kcal .
=> Energy required to melt 1000 g of ice at -10℃ ,,
By the general formulae = E = ms∆t = 1000 × 0.5 × 10 = 5 kcal .
Energy left for forming water = 51 - 5 = 46kcal.
Now ,, specefic latent heat of fusion of ice = 80cal/g , this 80 cal energy is required for fusion of 1 g ice .
=> 46000 cal energy will fuse x gram ice .
=> x = 46000÷80 = 575 g ice.
we know that the Density of water = 1 ( in CGS )
Hence ,, D = m/v = v = m/D = 575/1
➡ Volume of water available for drinking
= 575 ml
Hope It Helps You ^_^
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