Water is heated in an insulated, constant diameter tube by a 11-kw electric resistance heater. if the water enters the heater steadily at 15°c and leaves at 70°c, determine the mass flow rate of the water. consider the specific heat of water at room temperature is cp = 4.18 kj/kg·°c.

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

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.