wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Work done in converting one gram of ice at 10C into steam at 100C is

A
3045 J
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
6056 J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
721 J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
616 J
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 3045 J
Work done in converting 1gm of ice at 10C to steam at 100C
= Heat supplied to raise temperature of 1gm of ice from 10C to 0C [m×Cice×ΔT]
+ Heat supplied to convert 1 gm ice into water at 0C[m×Lice]
+ Heat supplied to raise temperature of 1gm of water from
0C to 100C[m×Cwater×ΔT]
+ Heat supplied to convert 1 gm water into steam at 100C[m×Lvapour]
=[m×Cice×ΔT]+[m×Lice]+[m×Cwater×ΔT]+[m×Lvapour]=[1×0.5×10]+[1×80]+[1×1×100]+[1×540]=725 calorie=725×4.2=3045 J


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Calorimetry
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon