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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoCalorimetryMedium2 minQB

PhysicsMediumsingle choice

$1 g$ of steam at $10 0^{\circ} C$ and an equal mass of ice at $0^{\circ} C$ are mixed. The temperature of the mixture in steady state will be: [Latent heat of steam = $540 cal$ g $^{- 1}$ and latent heat of ice = $80 cal/g$ ]

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Answer:

Solution:

Heat needed to turn the ice into water at $10 0^{\circ} C$ :
$Q_{1} = 1 \times 80 + 1 \times 1 \times 100 = 180 cal$
Heat given by the steam on condensing:
$Q_{2} = 1 \times 540 = 540 cal$
$Q_{2} > Q_{1}$
Steam condenses only partly, so the mixture stays at $10 0^{\circ} C$ .

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Calorimetry

⏱ 2mℹ️ Source: QB

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