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: $β = Q_{c} / W = T_{c} / (T_{h} - T_{c})$ .

How many calories of heat are discarded to the room ?

Options:

Answer:

Solution:

Water freezes at 0°C and releases $Q_{c} = m L = 1 \times 80 = 80 kcal$ .
For a Carnot refrigerator, $β = \frac{T _{c}}{T _{h} - T _{c}} = \frac{273}{300 - 273}$
$β = \frac{273}{27} \approx 10.1$ .
Input work $W = Q_{c} / β \approx 7.9 kcal$ .
Heat discarded to the room $Q_{h} = Q_{c} + W \approx 87.9 kcal$ , option B.

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

⏱ 3mℹ️ Source: QB

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