Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoCalorimetryMedium2 minQB

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Passage / Comprehension

In a container of negligible heat capacity, $200 gm$ ice at $0^{\circ} C$ and $100 gm$ steam at $10 0^{\circ} C$ are added to $200 gm$ of water that has temperature $5 5^{\circ} C$ . Assume no heat is lost to the surroundings and the pressure in the container is constant $1.0 atm$ . ( $L_{f} = 80$ cal/gm, $L_{v} = 540$ cal/gm, $s_{w} = 1$ cal/gm $^{\circ} C$ )

What is the final temperature of the system?

Options:

Answer:

Solution:

Heat needed to bring the ice and water to $10 0^{\circ} C$ :
$Q_{1} = 200 \times 80 + 200 \times 100 + 200 \times 45 = 45000 cal$
Heat released if all steam condenses at $10 0^{\circ} C$ :
$Q_{2} = 100 \times 540 = 54000 cal$
$Q_{2} > Q_{1}$
So the steam does not condense completely and the final temperature is $10 0^{\circ} C$ .

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Calorimetry

⏱ 2mℹ️ Source: QB

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