Hello i was wondering if someone could help me out with thisproblem. For part a

Question

Hello i was wondering if someone could help me out with thisproblem. For part a its just the sum of both temperatures rightsince no heat is lost?? 150 degrees celsius?
In a container of negligible mass, 0.0400 kg of steam at100°C and atmospheric pressure is added to 0.200 kg of water at 50.0°C. a) If no heat is lost to the surroundings, what is the finaltemperature of the system? b) At the final temperature, how many kilograms of steam andhow many kilograms of liquid water are there? Hello i was wondering if someone could help me out with thisproblem. For part a its just the sum of both temperatures rightsince no heat is lost?? 150 degrees celsius?
In a container of negligible mass, 0.0400 kg of steam at100°C and atmospheric pressure is added to 0.200 kg of water at 50.0°C. a) If no heat is lost to the surroundings, what is the finaltemperature of the system? b) At the final temperature, how many kilograms of steam andhow many kilograms of liquid water are there?

Explanation / Answer

It is a calorietry problem; a bit tricky. Steam at 100oC is in contact with water at50oC. Steam starts condensing losing(giving away) L = 2260J/g ofenergy This is absorbed by the water at 50oC and itstemperature rises. The rise in temp depends on the the amountof steam changing to water. If all of the steam become water(at 100oC), heatlost          Qs= m.L = 0.040 kg x 2260J/g = 90.4 kJ = 90,400 J ---(1) If all the water is heated up to 100oC,heat gained by it,          Qw= m.c. = 0.200 kgx4.18J/g.K x(100-50)oC               = 41.8 kJ = 41,800 J ------------------------------------(2) By inspection we note that to supply the heat Qwrequired for heating the water to 100oC, isonly aboutone half of the steam is required to condense. The actual amount of steam requitred to be condensed,                m' = Qw/L = 41800J/2260J/g = 18.5 g = 0.0185 kg. So, out of the available 0.040 kg of steam only 0.0185 kg ofsteam(at 100oC) condenses, raising the temp of thewater to 100oC. (a) The equilibrium temp is 100oC (b) The amount of steam left,             m'' = 0.040-0.0185 = 0.0215 kg = 21.5 g.      The amount of liquid water left,(original + condensed)              m1 =0.200 kg + 0.0185 = 0.2185 kg = 218.5 g Note: The final temp could be 100oC or less, notmore.          The maxamount of water left at the end= mass of steam + mass of water inthe beginning.                

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