5 g steam at 10 0 ∘ C is passed into 6 g ice at 0 ∘ C . Latent heats of steam and ice are 540 cal/g and 80 cal/g. Final temperature is:

Melting and heating the ice to 100°C needs Q = 6 ( 80 + 100 ) = 1080 cal . Only 540 1080 = 2 g steam must condense. 5 g is supplied, so the final mixture can remain at 100°C.

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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoThermal ExpansionMedium2 minQB

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5 g steam at $10 0^{\circ} C$ is passed into 6 g ice at $0^{\circ} C$ . Latent heats of steam and ice are 540 cal/g and 80 cal/g. Final temperature is:

Options:

Answer:

Solution:

Melting and heating the ice to 100°C needs

$Q = 6 (80 + 100) = 1080 cal .$

Only $\frac{1080}{540} = 2 g$ steam must condense. 5 g is supplied, so the final mixture can remain at 100°C.

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermal Expansion

⏱ 2mℹ️ Source: QB

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