JEE Main 2022 — Physics Thermodynamics

2022integermedium

A geyser heats water flowing at a rate of $2.0\mathrm{kg}$ per minute from $30^{\circ}C$ to $70^{\circ}C$ . If geyser operates on a gas burner, the rate of combustion of fuel will be _____ $g{\mathrm{min}}^{-1}$ .

[Heat of combustion $=8\times {10}^{3}{\mathrm{Jg}}^{-1}$ , Specific heat of water $=4.2{\mathrm{Jg}}^{-1}{C\circ }^{-1}$ ]

Official previous-year question

Held on 26 Jun 2022 · Verified 6 Jul 2026.

Did you get this right?

Solution

Given here, mass flow rate of water, $\frac{dm}{dt}=2\mathrm{kg}{\mathrm{min}}^{-1}$

The geyser heats the water, raising the temperature from ${\theta }_{1}=30^{\circ}C$ to ${\theta }_{2}=70^{\circ}C$ .

Total heat utilised in raising the temperature, $\frac{\Delta Q}{\Delta t}=\frac{dm}{dt}\times s\Delta \theta$

$\frac{\Delta Q}{\Delta t}=\frac{2}{60}\times 4.2\times 40=\frac{m}{t}\times L$

$L=$ heat of combination $=8\times {10}^{3}J{g}^{-1}$

Now, rate of consumption of fuel is $\frac{m}{t}=\frac{2\times 4.2\times 40}{60\times 8\times {10}^{3}}$

$=\frac{4.2}{6}\times {10}^{-3}=0.7\times {10}^{-3}\mathrm{kg}{\mathrm{sec}}^{-1}$

$=42g{\mathrm{min}}^{-1}$

Your note

JEE Main Thermodynamics in other years

Thermodynamics 2026 Thermodynamics 2025 Thermodynamics 2024 Thermodynamics 2023 Thermodynamics 2021 Thermodynamics 2020 Thermodynamics 2019 Thermodynamics 2018 Thermodynamics 2017 Thermodynamics 2016 Thermodynamics 2015 Thermodynamics 2014

Keep practicing Thermodynamics

Work through every JEE Main Thermodynamics PYQ, year by year.

Browse all Physics topics

Solution

Given here, mass flow rate of water, $\frac{dm}{dt}=2\mathrm{kg}{\mathrm{min}}^{-1}$

The geyser heats the water, raising the temperature from ${\theta }_{1}=30^{\circ}C$ to ${\theta }_{2}=70^{\circ}C$ .

Total heat utilised in raising the temperature, $\frac{\Delta Q}{\Delta t}=\frac{dm}{dt}\times s\Delta \theta$

$\frac{\Delta Q}{\Delta t}=\frac{2}{60}\times 4.2\times 40=\frac{m}{t}\times L$

$L=$ heat of combination $=8\times {10}^{3}J{g}^{-1}$

Now, rate of consumption of fuel is $\frac{m}{t}=\frac{2\times 4.2\times 40}{60\times 8\times {10}^{3}}$

$=\frac{4.2}{6}\times {10}^{-3}=0.7\times {10}^{-3}\mathrm{kg}{\mathrm{sec}}^{-1}$

$=42g{\mathrm{min}}^{-1}$

JEE Main 2022 — Physics Thermodynamics

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics

JEE Main 2022 — Physics Thermodynamics

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics