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Heat and Thermodynamics question

2023 · 31 Jan · Shift 2 · Q63
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  5. /2023 · 31 Jan · Shift 2 · Q63

Heat and Thermodynamics question

2023 · 31 Jan · Shift 2 · Q63

JEE MainPhysicsHeat and ThermodynamicsNumerical+4 / −1
A water heater of power 2000 W2000 \mathrm{~W}2000 W is used to heat water. The specific heat capacity of water is 4200 Jkg−1 K−14200 \mathrm{~J}\mathrm{kg}^{-1} \mathrm{~K}^{-1}4200 Jkg−1 K−1. The efficiency of heater is 70%70 \%70%. Time required to heat 2 kg2 \mathrm{~kg}2 kg of water from 10∘C10^{\circ} \mathrm{C}10∘C to 60∘C60^{\circ} \mathrm{C}60∘C is ‾\underline{\hspace{2cm}}​ s. (Assume that the specific heat capacity of water remains constant over the temperature range of the water).
Numerical answer
View written solutionFree

Correct answer: 300

  1. Given data

    • Power of heater: P=2000 WP = 2000\,\text{W}P=2000W
    • Efficiency: η=70%=0.7\eta = 70\% = 0.7η=70%=0.7
    • Mass of water: m=2 kgm = 2\,\text{kg}m=2kg
    • Specific heat capacity of water: c=4200 J kg−1K−1c = 4200\,\text{J kg}^{-1}\text{K}^{-1}c=4200J kg−1K−1
    • Initial temperature: 10∘C10^\circ\text{C}10∘C
    • Final temperature: 60∘C60^\circ\text{C}60∘C

  2. Temperature rise ΔT=60−10=50 K\Delta T = 60 - 10 = 50\,\text{K}ΔT=60−10=50K

  3. Heat required by water Q=mcΔTQ = mc\Delta TQ=mcΔT Q=2×4200×50Q = 2 \times 4200 \times 50Q=2×4200×50 Q=420000 JQ = 420000\,\text{J}Q=420000J

  4. Useful power delivered to water Since heater efficiency is 70%70\%70%, effective power is Puseful=ηP=0.7×2000=1400 WP_{\text{useful}} = \eta P = 0.7 \times 2000 = 1400\,\text{W}Puseful​=ηP=0.7×2000=1400W

  5. Time required t=QPusefult = \frac{Q}{P_{\text{useful}}}t=Puseful​Q​ t=4200001400=300 st = \frac{420000}{1400} = 300\,\text{s}t=1400420000​=300s

  6. Final answer 300\boxed{300}300​

The derived answer matches the stored correct answer.

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