We have half a bucket (6 ℓ ) of water at 2 0 ∘ C . If we want water at 4 0 ∘ C , how much steam at 10 0 ∘ C should be added to it?

Take L v = 540 cal/g. Heat gained by water = heat lost by steam (condensing, then cooling to 4 0 ∘ C ): 6000 × 1 × ( 40 − 20 ) = m × 540 + m × 1 × ( 100 − 40 ) 12 × 1 0 4 = 600 m m = 200 g

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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoCalorimetryMedium2 minQB

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We have half a bucket (6 $ℓ$ ) of water at $2 0^{\circ} C$ . If we want water at $4 0^{\circ} C$ , how much steam at $10 0^{\circ} C$ should be added to it?

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Answer:

Solution:

Take $L_{v} = 540$ cal/g. Heat gained by water = heat lost by steam (condensing, then cooling to $4 0^{\circ} C$ ):
$6000 \times 1 \times (40 - 20) = m \times 540 + m \times 1 \times (100 - 40)$
$12 \times 1 0^{4} = 600 m$
$m = 200 g$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Calorimetry

⏱ 2mℹ️ Source: QB

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