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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Thermochemistry measures energy changes in chemical reactions and distinguishes exothermic from endothermic processes. Work covers systems, surroundings and state functions, internal energy, the first law of thermodynamics, enthalpy and Hess's law, including the sign conventions used when heat enters or leaves a system.
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