A geyser, operating on LPG (liquefied petroleum gas) heats water flowing at the rate of 3.0 litres p — Thermodynamics and Thermochemistry Chemistry Question
Question
A geyser, operating on LPG (liquefied petroleum gas) heats water flowing at the rate of 3.0 litres per minutes, from $27^\circ\text{C}$ to $77^\circ\text{C}$. If the heat of combustion of LPG is 40,000 J/g, how much fuel, in g, is consumed per minute? (Specific heat capacity of water is 4200 J/kg-K)
Answer: C
💡 Solution & Explanation
Mass of water per minute = $3.0\text{ kg} = 3000\text{ g}$. Temperature rise = $77 - 27 = 50\text{ K}$. Heat required per minute $Q = mc\Delta T = 3.0\text{ kg} \times 4200\text{ J/kg-K} \times 50\text{ K} = 630,000\text{ J}$. Fuel consumed = $630,000\text{ J} / 40,000\text{ J/g} = 15.75\text{ g}$.
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