An electric appliance supplies 6000 {J} / {min} heat to the system. If the system delivers a power of 90 {W}. How long it would take to increase the internal energy by 2.5 × 10³ {J} ? [26 Aug 2021 Shift 1]

Correct Answer is : 2.5×102s2.5 × 10² {s}2.5×102s

Correct Answer is : 2.5×102s2.5 × 10² {s}2.5×102s

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Thermodynamics Part 3

Section: Physics

Question : 77 of 100

Marks: +1, -0

6000 \mathrm{J} / \mathrm{min}

90 \mathrm{W}

2.5 \times 10^{3} \mathrm{J}

Solution:

Given, heat supplied to the system,

\frac{\Delta Q}{\Delta t}=6000 \mathrm{J} / \mathrm{min}=\frac{6000}{60} \mathrm{J} / \mathrm{s}=100 \mathrm{J} / \mathrm{s}

Power delivered,

P=\frac{\Delta W}{t}=90 \mathrm{W}

Increase in internal energy,

\Delta U=2.5 \times 10^{3} \mathrm{J}

From first law of thermodynamics, we have

\Delta Q=\Delta U+\Delta W

\frac{\Delta Q}{\Delta t}=\frac{\Delta U}{\Delta t}+\frac{\Delta W}{\Delta t}

...(i)

Substituting the given values in Eq. (i), we get

100=\frac{2.5 \times 10^{3}}{\Delta t}+90

\Rightarrow 10=\frac{2.5 \times 10^{3}}{\Delta t}

\Rightarrow \Delta t=\frac{2.5 \times 10^{3}}{10}

⇒

\Delta t=2.5 \times 10^{2} \mathrm{s}

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