Question

How many calories are required to convert 2 grams of ice at 0 degrees C to water vapor at 100 degrees C?

Answer 1

To convert 2 grams of ice at 0 degrees C to water vapor at 100 degrees C, 1439.8 calories are required. This includes the heat of fusion (159.8 calories), the heat to raise the temperature to 100 C (200 calories), and the heat of vaporization (1080 calories).

Step-by-step explanation:

To calculate how many calories are required to convert 2 grams of ice at 0 degrees C to water vapor at 100 degrees C, we need to consider the following steps:

• Melting the ice into liquid water, which requires heat known as the heat of fusion.
• Heating the liquid water from 0 degrees C to 100 degrees C.
• Converting the water at 100 degrees C into water vapor, which requires heat known as the heat of vaporization.

Using the heat of fusion for water (79.9 cal/g) and the heat of vaporization of water (540 cal/g at 100 degrees C), along with the specific heat capacity for liquid water (1 cal/g°C), we can perform the calculations:

1. Heat to melt ice: Heat of fusion × Mass = 79.9 cal/g × 2 g = 159.8 calories.
2. Heat to raise temperature to 100°C: Specific heat capacity × Mass × Temperature change = 1 cal/g°C × 2 g × (100°C - 0°C) = 200 calories.
3. Heat to vaporize water: Heat of vaporization × Mass = 540 cal/g × 2 g = 1080 calories.

Add all three quantities to get the total number of calories required:

159.8 calories + 200 calories + 1080 calories = 1439.8 calories

Therefore, 1439.8 calories are required to convert 2 grams of ice at 0 degrees C to water vapor at 100 degrees C.

Answer 2

360.8987 GRAM CALORIE OR 0.3608987 KILOCALORIE IS THE CALORIE REQUIRED TO CONVERT 2 GRAMS OF ICE AT 0 DEGREES C TO WATER VAPOUR AT 100 DEGREES C.

Step-by-step explanation:

To convert 2 g of ice at 0 C to water vapor at 100 C:

First, the ice will be converted to water at 0 C

Then, water at 0 C to water vapor at 100 C

We must know that the latent heat of fusion for ice is 3.35*10^5 J/kg

the standard heat of enthalpy of water = 4200 J/kg

So therefore

Total heat to convert ice at 0 c to water vapor at 100 c is:

= M Lf + mc Q

= 2/ 1000 * 3.35 * 10^5 + 2/1000 * 4200 * (100 -0)

= 0.002 * 3.35 ^ 10^5 + 0.002 * 4200 * 100

= 0.0067 * 10 ^5 + 840

= 670 + 840

= 1510 J

The heat required to convert 2 g of ice at 0 c to water vapor at 100 C is 1510 J and to get the calorie value we divide the answer in joules by 4.184.

1510 joules = 360.8987 gram calories or 0.3608987 kilocalorie.