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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoSecond Law of ThermodynamicsMedium2 minPYQ_2016

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A refrigerator works between $4^{o} C$ and $30^{o} C$ . It is required to remove $600$ calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is:

(Take $1$ cal = $4 .2$ Joules)NEET 2016 (Phase 1)

Options:

Answer:

Solution:

$\frac{Q_{2}}{Q_{1}} = \frac{T_{2}}{T_{1}} \Rightarrow \frac{600 + w}{600} = \frac{303}{277} = \frac{277 + 26}{277}$

$1 + \frac{w}{600} = 1 + \frac{26}{277}$

$w = 600 \times \frac{26}{277} \times 4.2$

$w = 236.5$

Stream:NEETSubject:PhysicsTopic:Heat & ThermoSubtopic:Second Law of Thermodynamics

⏱ 2mℹ️ Source: PYQ_2016

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