Question

How much heat is required to convert 8.0 g of ice at -15^@ to steam at 100^@? (Given, c_(ice) = 0.53 cal//g-^@C, L_f = 80 cal//g and L_v = 539 cal//g, and c_(water) = 1 cal//g-^@C) .

Answer 1

Hence total heat required = 5,815.6 J / 4.184 = 1389 calories

Explanation:

First we have to bring the ice to 0 C heat required  = 8 × 15 × 0.53 J = 63.6 J

Heat required to melt 8 g of ice = 8 × 539 J = 4,312 J

Heat required for 8 g water from 0 C to 100 C= 8 ×100× 1 J = 800 J

Heat required to change it to steam = 80 × 8 J = 640 J

Total heat required = 63.6 + 4,312 + 800 + 640 = 5,815.6 J  

Now convert Joule into calorie by dividing with 4.184

Total heat required = 5,815.6 J / 4.184 = 1389 calories