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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoSecond Law of ThermodynamicsMedium3 minQB

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Passage / Comprehension

A Carnot refrigerator absorbs heat from water at 0°C and releases it into a room at 27°C.
It turns 1 kg of water at 0°C into ice at 0°C. The latent heat of ice is 80 kcal/kg.
The coefficient of performance is heat taken from the cold body divided by the work input: $β = Q_{c} / W = T_{c} / (T_{h} - T_{c})$ .

What is the work done by the refrigerator in this process ( 1 cal = 4.2 joule)

Options:

Answer:

Solution:

With $Q_{c} = 80 kcal$ and $Q_{h} \approx 87.9 kcal$ , input work is
$W = Q_{h} - Q_{c} = 7.9 kcal$ .
Converting units, $W = 7.9 \times 4.2 = 33.18 kJ$ , option B.

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

⏱ 3mℹ️ Source: QB

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