Mix 20 g ice at -10°C with 10 g water at 40°C in an insulated vessel. Use ci = 0.5 , cw = 1 cal / ( g ∘ C ) and Lf = 80 cal / g . Find ice remaining in g to two decimals.

Water supplies 400 cal. Warming ice uses 20 × 0.5 × 10 = 100 cal. Remaining 300 cal melts 3.75 g. Ice left = 20 − 3.75 = 16.25 g at 0 ∘ C .

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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoThermal ExpansionMedium2 minQB

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Mix 20 g ice at -10°C with 10 g water at 40°C in an insulated vessel. Use $ci = 0.5, cw = 1 cal / (g^{\circ} C)$ and $Lf = 80 cal / g$ . Find ice remaining in g to two decimals.

Answer:

16.25

Solution:

Water supplies 400 cal. Warming ice uses $20 \times 0.5 \times 10 = 100$ cal. Remaining 300 cal melts 3.75 g. Ice left $= 20 - 3.75 = 16.25 g$ at $0^{\circ} C$ .

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermal Expansion

⏱ 2mℹ️ Source: QB

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