Mix 100 g water at 80°C with 50 g ice at 0°C. No losses or vessel heat. cw = 1 cal / ( g ∘ C ) , Lf = 80 cal / g . Find final °C rounded to nearest integer.

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Mix 100 g water at 80°C with 50 g ice at 0°C. No losses or vessel heat. $cw = 1 cal / (g^{\circ} C), Lf = 80 cal / g$ . Find final °C rounded to nearest integer.

Answer:

Solution:

Available heat to $0^{\circ} C = 8000 cal > 4000 cal$ melting heat. Then $100 (80 - T) = 4000 + 50 T$ , yielding $T =$ 26.6667°C, rounding to 27.

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermal Expansion

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