A cylinder of gas is assumed to contain 14 kg of butane. A normal family requires 20,000 kJ of energy per day for cooking. How many days will the butane gas in the cylinder last?
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Answered by
17
Heya
here's your answer...
Molar mass of butane, C4H10=12×4+10=58gmol−1
58 g of butane gives 2658 kJ of heat energy.
∴ 14 kg of butane will give heat energy = 2658kJ×(14×103g)58g=641.5862×103kJ
Daily energy requirement for cooking = 20,000 kJ = 2×104 kJ day−1
∴ Number of days cylinder will last = 641.5862×103 kJ 2×104 kJ day−1 = 32.08 days
hope it helps....✌✌
here's your answer...
Molar mass of butane, C4H10=12×4+10=58gmol−1
58 g of butane gives 2658 kJ of heat energy.
∴ 14 kg of butane will give heat energy = 2658kJ×(14×103g)58g=641.5862×103kJ
Daily energy requirement for cooking = 20,000 kJ = 2×104 kJ day−1
∴ Number of days cylinder will last = 641.5862×103 kJ 2×104 kJ day−1 = 32.08 days
hope it helps....✌✌
Answered by
6
Answer:32 days
Explanation:
Molar mass of butane,
C4H10=12×4+10=58g/mol
58 g of butane gives 2658 kJ of heat energy.
∴ 14 kg of butane will give heat energy = 2658kJ×(14×103g)58g=641.5862×10kJ
Daily energy requirement for cooking = 20,000 kJ = 2×104 kJ/day
∴ Number of days cylinder will last = 641.5862×103 kJ 2×10^4 kJ/mol
=32.08 days
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