heat required to convert 1 gram of ice at zero degree Celsius into steam at hundred degree Celsius ( given L steam = 536 cal/gm)
Answers
Answered by
6
Answer:
Mass of ice=1g
Temp of ice=0 degree C
1 First the ice at 0 degree C is to be converted into water at 0 degree C
Heat required at this stage
= Mass of the ice x Latent heat of fusion of ice
=1x 80 =80 cal
2 Now the temperature of water is to be increased from 0 to 100 degree C
Heat required for this =Mass of water x rise in temperature x specific heat of water
=1 x100 x1= 100 cal
3 Now this water at100 degree C is to be converted into vapour
Heat required for this
=Mass of water x Latent heat
= 1 x 536 =536 cal
Total heat required
=80 +100 +536=716 cal
MARK ME AS BRAINLIEST IF YOU FIND THIS HELPFUL :)
Similar questions
English,
3 months ago
Social Sciences,
3 months ago
Biology,
7 months ago
Math,
10 months ago
Science,
10 months ago
Social Sciences,
10 months ago