Flag Thermal Physics> calorimetry...
question mark

1 kg of ice at 00C is mixed with 1 kg of steam at 1000C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fussion of ice=3.36*105 J/kg and latent heat of vaporization of water=2.26*106 J/kg.

pranshu aggarwal , 14 Years ago
Grade
anser 3 Answers
Babulal Prasad

Last Activity: 14 Years ago

heat needed to melt the ice = mL = 1* 3.36*100000 =336000 J heat needed to increase the temperature of water to 100 degree centigrade=1*4200=4200 J total heat requirement= 340200 suppose m mass of steam is converted in to water then m*2260000= 340200 m= 340200/2260000=150.5 g final composition water=1150.5 g & steam=849.5 g

Vidya Iyer

Last Activity: 6 Years ago

Energy required to melt ice completely= 1000*80 cal
Energy required to raise it's temperature to 100°= 1000*1*100
Total energy=1000(80+100) cal
Mass of steam required= m
m*540*1000 =180*1000
m= 0.33kg

Kushagra Madhukar

Last Activity: 4 Years ago

Dear student,
Please find the solution to your problem below.
 
Heat needed to melt the ice = mL = 1* 3.36*100000 =336000 J
Heat needed to increase the temperature of water to 100 degree centigrade=1*4200=4200 J
Total heat requirement= 340200 J
Suppose m mass of steam is converted in to water
then, m*2260000= 340200
m = 340200/2260000 = 150.5 g
Final composition water = 1150.5 g & steam=849.5 g
 
Thanks and regards,
Kushagra

star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments