1 g of ice at 0oC is converted into steam at 100oC. The amount of heat required to do so is
A
80 cal
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
536 cal
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
616 cal
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.716 kcal
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D 0.716 kcal Heatrequired=(heatrequiredtoconverticeat0oCto100oC)+(Heatrequiredtoconvertwatertosteamat100oC) =(1×80cal)+(1×1×100)+(1×536)=(180+536) =716cal.or0.716kcal.