According to the thermal formula Q=vq=cm water △t (assuming no heat loss during heat transfer)
The calorific value of the gas natural q1=3.14×10^7J/m ^3, The volume of natural gas v=1m^3, the specific heat capacity of water c=4.2×10?J/(㎏·℃) Suppose let the normal temperature be 20°C, the local atmospheric pressure is a standard atmospheric pressure, so the temperature change △t=(100-20 )°C=80°C Put this data into the formula to obtain, m of water is approximately 93.45 kg. In real situations, heat is lost and therefore the quality of boiled water is lower.
1 kilowatt hour of electricity = 1kw h = 3.6 × 10^6J = Q = cm of water △t
By entering the data, m of water is equivalent to about 10.71 kg In. real conditions, the amount of heat It is lost again, therefore the quality of boiled water is less.
The above data are theoretical values and are for reference only.reference only.
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