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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 $^{\circ} C$ )

At the final temperature, mass of the total water present in the system, is

Options:

Answer:

Solution:

Let $x$ gm of steam condense. Heat released by steam equals heat absorbed:
$540 x = 45000$
$540 x = 200 \times 80 + 200 \times 100 + 200 \times 45$
$x = \frac{250}{3} = 83.33 g$
Total water = initial water + ice + condensed steam:
$m = 400 + 83.33 = 483.3 gm$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Calorimetry

⏱ 2mℹ️ Source: QB

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