Respuesta :
Answer:
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.
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.
Amount calories are required to convert into vapor is 1,440 calories
Latent heat of water:
Given that;
Amount of ice = 2 grams
Latent heat of water = 540 calorie
Find:
Amount calories are required to convert into vapor
Computation;
Heat required to melt at 0°C to water at 0°C = 2 × 80
Heat required to melt at 0°C to water at 0°C = 160 calorie
Water from 0°C to 100°C = 2 × 100
Water from 0°C to 100°C = 200 calorie
Water from 100°C to 100°C vapor = 2 × 540
Water from 100°C to 100°C vapor = 1,080 calorie
Amount calories are required to convert into vapor = 160 + 200 + 1080
Amount calories are required to convert into vapor = 1,440 calories
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