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Heat and Thermodynamics question

2022 · 26 Jun · Shift 2 · Q67
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  5. /2022 · 26 Jun · Shift 2 · Q67

Heat and Thermodynamics question

2022 · 26 Jun · Shift 2 · Q67

JEE MainPhysicsHeat and ThermodynamicsNumerical+4 / −1
A geyser heats water flowing at a rate of 2.0 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 ‾\underline{\hspace{2cm}}​ g min −-− 1. [Heat of combustion = 8 ×\times× 103 Jg −-− 1, Specific heat of water = 4.2 Jg −-− 1 ∘^\circ∘ C −-− 1]
Numerical answer
View written solutionFree

Correct answer: 42

  1. Given data

    • Mass flow rate of water =2.0 kg min−1=2000 g min−1= 2.0\ \text{kg min}^{-1} = 2000\ \text{g min}^{-1}=2.0 kg min−1=2000 g min−1
    • Initial temperature =30∘C= 30^\circ \text{C}=30∘C
    • Final temperature =70∘C= 70^\circ \text{C}=70∘C
    • Rise in temperature: ΔT=70−30=40∘C\Delta T = 70 - 30 = 40^\circ \text{C}ΔT=70−30=40∘C
    • Specific heat of water: c=4.2 J g−1 ∘C−1c = 4.2\ \text{J g}^{-1}\ ^\circ\text{C}^{-1}c=4.2 J g−1 ∘C−1
    • Heat of combustion of fuel: H=8×103 J g−1H = 8 \times 10^3\ \text{J g}^{-1}H=8×103 J g−1

  2. Heat required per minute to heat the water

    Using, Q=mcΔTQ = mc\Delta TQ=mcΔT

    Substitute the values: Q=2000×4.2×40Q = 2000 \times 4.2 \times 40Q=2000×4.2×40

    Q=336000 J min−1Q = 336000\ \text{J min}^{-1}Q=336000 J min−1

  3. Fuel required per minute

    If xxx g of fuel burns per minute, then heat produced is: x×8×103 J min−1x \times 8 \times 10^3\ \text{J min}^{-1}x×8×103 J min−1

    Equating this to required heat: x×8×103=336000x \times 8 \times 10^3 = 336000x×8×103=336000

    x=3360008000=42x = \frac{336000}{8000} = 42x=8000336000​=42

  4. Final answer 42 g min−1\boxed{42\ \text{g min}^{-1}}42 g min−1​

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