- #1
vikmak
- 4
- 0
Is the below calc right?
Input electrical power for heating- 4KW
Heating time= 20 hrs = 72,000 sec
Input energy available= = 288,000,000 J
Qty of water to be heated- 7,500Kg
Specific heat of water- 4,200 joules/Kg/deg C
Delta T in Celsius that can be achieved with the said input energy is 9.14 ?
Input electrical power for heating- 4KW
Heating time= 20 hrs = 72,000 sec
Input energy available= = 288,000,000 J
Qty of water to be heated- 7,500Kg
Specific heat of water- 4,200 joules/Kg/deg C
Delta T in Celsius that can be achieved with the said input energy is 9.14 ?