Answer:
[tex]\dot m = 0.047 kg/s[/tex]
Explanation:
we know by energy balanced for steady flow system that control volume W is given as
[tex]W = \dot m c_p( T_{OUT} - T_{IN})[/tex]
here
[tex]\dot m[/tex] is mass flow rate
we have W = 11 kW
Outer temperature = 70°C
Inner temperature = 15°C
therefore we have mass flow rate
[tex]\dot m = \frac{W}{c_p *(T_{OUT} - T_{IN})}[/tex][tex]= \frac{11}{4.18 *(70-15)}[/tex]
[tex]\dot m = 0.047 kg/s[/tex]
Answer:
The mass flow rate of water is 0.0478 kg/s or 47.8 g/s
Explanation:
We have:
Power of electric heater = P = 11 KW
Change in temperature of water = ΔT = 70°C - 15°C = 55°C
Specific heat of water = C = 4.18 KJ / kg.°C
Now, from conservation of energy, we know that:
Energy dissipated by heater = Energy acquired by Water
Energy dissipated by heater = m C ΔT
Dividing both sides by time (t), we get:
Power of Heater = (mass flow rate) C ΔT
Mass Flow Rate = P/CΔT
Mass Flow Rate = (11 KW)/(4.18 KJ/kg.°C)(55°C)
Mass Flow Rate = 0.0478 kg/s = 47.8 g/s
