How many gram of butane must be burnt to heat 10 kg water from 30°C to 100°C. Heat of combustion of butane is −700 kcal/mol and specific heat of water is 1 cal gm−1 K−1 .
Correct answer: A. 58 gm
- A. 58 gm
- B. 56 gm
- C. 54 gm
- D. 52 gm
Explanation
First, let's calculate the heat required to raise the temperature of water from 30°C to 100°C: Heat (Q) = mass (m) × specific heat (c) × change in temperature (ΔT) Q = 10000 g × 1 cal/g°C × (100°C - 30°C) Q = 700000 cal Now, let's convert the heat into kcal: 700000 cal ÷ 1000 = 700 kcal Now, let's calculate the moles of butane required to produce 700 kcal of heat: Moles = Heat ÷ Heat of Combustion Moles = 700 kcal ÷ -700 kcal/mol (note: -700 kcal/mol is the same as 700 kcal/mol in magnitude) Moles = 1 mol The molar mass of butane (C4H10) is approximately 58 g/mol. So, the mass of butane required = moles × molar mass = 1 mol × 58 g/mol = 58 g.
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