How many calories are required to change one gram of 0$\degree$C ice to 100$\degree$C steam? The latent heat of fusion is 80 cal/g and the latent heat of vaporization is 540 cal/g. The specific heat of water is 1.00 cal/g . K.

Question

Quizplus · Accepted Answer

To change ice at 0°C to water at 0°C, we need to add heat = (mass of ice) x (latent heat of fusion) = 1g x 80cal/g = 80cal.
To raise the temperature of 1g of water from 0°C to 100°C, we need to add heat = (mass of water) x (specific heat capacity of water) x (change in temperature) = 1g x 1cal/g°C x 100°C = 100cal.
To change water at 100°C to steam at 100°C, we need to add heat = (mass of water) x (latent heat of vaporization) = 1g x 540cal/g = 540cal.
Total heat required = 80cal + 100cal + 540cal = 720cal. 
Therefore, the answer is D.

How Many Calories Are Required to Change One Gram of 0