A calorimeter of heat capacity 100 cal/°C holds 200 g water and 50 g ice at 0°C. Add 100°C steam until 20°C equilibrium. cw = 1 cal / ( g ∘ C ) , Lf = 80 , Lv = 540 cal / g . Steam mass in g , nearest

Question

A calorimeter of heat capacity 100 cal/°C holds 200 g water and 50 g ice at 0°C. Add 100°C steam until 20°C equilibrium. cw = 1 cal / ( g ∘ C ) , Lf = 80 , Lv = 540 cal / g . Steam mass in g , nearest integer?

TestHub · Accepted Answer

Heat needed = 50 ( 80 ) + 250 ( 20 ) + 100 ( 20 ) = 11000 cal . Steam supplies 540 + 80 = 620 cal / g . Mass = 11000/620 = 17.7419 g , rounding to 18.

Physics - Heat & Thermo Question with Solution | TestHub

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