The amount of heat required to convert 1 gm of ice at (minus)100 degree Celsius to steam at 100 degree Celsius is:

A. 766 cal

B. 696 cal

C. 716 cal

D. 736 cal

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

1. =80 +100 +536=716 cal