5 g steam at 100°C is mixed with 10 g ice at 0°C. c w = 1 cal / ( g ∘ C ) , L f = 80 , L v = 540 cal / g . Select correct statements.

Ice to water at 10 0 ∘ C needs 10 ( 80 + 100 ) = 1800 cal . Condensed steam = 540 1800 = 3 10 g , less than 5 g. Water = 10 + 3 10 = 3 40 g ; steam = 5 − 3 10 = 3 5 g . Coexistence at 100°C gives B,C,D.

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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoThermal ExpansionMedium2 minQB

PhysicsMediummultiple choice

5 g steam at 100°C is mixed with 10 g ice at 0°C. $c_{w} = 1 cal / (g^{\circ} C), L_{f} = 80, L_{v} = 540 cal / g$ . Select correct statements.

Options:(select one or more)

Answer:

B, C, D

Solution:

Ice to water at $10 0^{\circ} C$ needs $10 (80 + 100) = 1800 cal$ . Condensed steam $= \frac{1800}{540} = \frac{10}{3} g$ , less than 5 g. Water $= 10 + \frac{10}{3} = \frac{40}{3} g$ ; steam $= 5 - \frac{10}{3} = \frac{5}{3} g$ . Coexistence at 100°C gives B,C,D.

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermal Expansion

⏱ 2mℹ️ Source: QB

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