how much heat is required to vaporize 7 grams of ice initially at 0 degrees celsius when the latent fusion of ice is 80 cal/g, the vaporization of water is 540 cal/g, and the specific heat of water is 1 cal/(g x C)?

Each gram of 0 deg. C ice requires 80 Cal to melt, 100 Cal to heat from 0 to 100 C as liquid water, and 540 cal to vaporize. That is a total of 720 calories.

Multiply that by the number of grams for the answer.

To calculate the heat required to vaporize 7 grams of ice initially at 0 degrees Celsius, we need to consider the different stages of the phase change: from solid ice to liquid water (latent fusion) and from liquid water to water vapor (vaporization).

1. First, let's calculate the heat required for the ice to reach its melting point:
The latent fusion (heat of fusion) of ice is given as 80 cal/g. Thus, the heat required to melt 7 grams of ice is:
Heat for melting = 80 cal/g x 7 g = 560 cal

2. Next, let's calculate the heat required to raise the temperature of the resulting water from 0 degrees Celsius to its boiling point (100 degrees Celsius):
The specific heat of water is given as 1 cal/(g x C). Therefore, the heat required to raise the temperature of 7 grams of water by 100 degrees Celsius is:
Heat for temperature change = 1 cal/(g x C) x 7 g x 100 C = 700 cal

3. Finally, let's calculate the heat required for the water to vaporize:
The heat of vaporization of water is given as 540 cal/g. Therefore, the heat required to vaporize 7 grams of water is:
Heat for vaporization = 540 cal/g x 7 g = 3780 cal

To find the total heat required, we add up the heat required for each stage:
Total heat required = Heat for melting + Heat for temperature change + Heat for vaporization
= 560 cal + 700 cal + 3780 cal
= 5040 cal

Therefore, 5040 calories of heat are required to vaporize 7 grams of ice initially at 0 degrees Celsius.