A gas based geyser heats water flowing at the rate of 5.0 litres per minute from 27^ C to 87^ C . The rate of consumption of the gas is \_\_\_\_ g / s . (Take heat of combustion of gas =5.0 × 10^ 4 ~J / g ) specific

Correct answer: A. 0.42 Flow rate = 5.0 L/min = 5.0 kg/min, Δ T = 87 - 27 = 60 °C Heat required per minute: Q = mcΔ T = 5.0 × 4200 × 60 = 1.26 × 10^6 J/min If m grams of gas consumed per minute: m × 5.0 × 10^4 = 1.26 × 10^6 m = (1.26 × 10^6)/(5.0 × 10^4) = 25.2 g/min Rate = (25.2)/(60) = 0.42 g/s

JEE Main 202621 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 21 January
Shift: Morning Shift
Subject: Physics
Chapter: Thermal Properties of Matter

JEE Main 2026 · 21 January · Morning Shift · Physics · Thermal Properties of Matter

A gas based geyser heats water flowing at the rate of 5.0 litres...

A gas based geyser heats water flowing at the rate of 5.0 litres per minute from

27^{\circ} \mathrm{C}

87^{\circ} \mathrm{C}

. The rate of consumption of the gas is

\_\_\_\_

\mathrm{g} / \mathrm{s}

.
(Take heat of combustion of gas

=5.0 \times 10^{4} \mathrm{~J} / \mathrm{g}

) specific heat capacity of water

=4200 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}

Options

A0.42
B2.1
C0.21
D4.2

Correct answer

A. 0.42

Step-by-step solution

Flow rate = 5.0 L/min = 5.0 kg/min,

\Delta T = 87 - 27 = 60

°C
Heat required per minute:

Q = mc\Delta T = 5.0 \times 4200 \times 60 = 1.26 \times 10^6

J/min
If m grams of gas consumed per minute:

m \times 5.0 \times 10^4 = 1.26 \times 10^6

m = \frac{1.26 \times 10^6}{5.0 \times 10^4} = 25.2

g/min
Rate

= \frac{25.2}{60} = 0.42

g/s

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