Verified answer

heat required Q=mc(t2-t1)

where c=specific heat capacity for ice =.5 calorie/kg

m=mass

t2= final temperature

t1=initial temperature

therefore Q=100*.5*(-.5+20)= 975 calories

Q=m*c*delta t

c of ice=.5 calories/gram

Q= 100*.5*15

Q = 750 calories

Specific heat of ice +0.5 cal/g C.

So, according to definition ,0.5 cal is required to raise the temperature of 1 g of ice by 1 degree centigrade.

Increase in temperature required -.5-(-20)=19.5 degree centigrade.

so, heat required for 1 g = 0.5 X19.5 cal

Heat required for 100 g = 0.5 X19.5 .X 100= 975 cal.