Calculation of Kcal

 Law of conservation of Energy

In physics, the law of conservation of energy states that the total energy of an isolated system remains constant.

Calculation of Kcal

kcal=1L water heating up 1°C
Ex:
  • 100L of 30°C water heating up to 32°C,need 200kcal
(32°C -30°C) * 100L=200kcal
  • 100L of 50°C water chilling to 40°C,need –1000kcal
(50°C-40°C)* 100L = 1000 kcal
It needs 80 kcal to make phase change of water and ice.
0°C water → 0°C ice = -80 kcal
0°C ice → 0°C water = 80 kcal
Ex:
How much cal. do 1 1°C water become to -2°C ice
(1°C-0°C) *1 kcal=1 kcal
(0°C-(-2°C))*0.5 kcal=1 kcal (When the water change into ice, it needs 0.5 kcal every 1°C of cooling)
Water→Ice = 80kcal
C Water → -2°C Ice = 1+1+80=82kcal
Phace change of water and steam
100°C Water 100°C Steam594 kcal
Heating efficiency of steam is great. If we heating up 100L water by 1L steam, the temp. will rise 5.9°C.

Other calculation of Kcal

  1. 1RT=3024 kcal (Metric Units) = 12000BTU(Imperial units)
  2. 1KW electricity = 860 kcal (860 in theory, the actual value is 650~700 kcal) →Lower efficiency of heating, higher cost.
  3. 1kg gas(heavy oil)=12000 kcal