Question

A continu a continuous flow water heater has an electrical power rating 2 kilowatt and efficiency of conversion of electrical power into heat is 80% the water is flowing through the device at the rate 100 CC per second and the inlet temperature is 10 degrees outlet temperature will be

Answer 1

Given Electrical Power = 2 kilowatt = 2000 w

The efficiency of the heater is 80%

We know that The Heat energy in joules (J) is equal to the Electric power P in watts (W), times the time period t in seconds (s):

In unit time Heat energy = Electric Energy

So the Heat energy absorbed by the water is $E_{P}=2000\times\frac{80}{100}=1600$J

We know that,

$\text{Heat Energy}=\text{mass}\times\text{specific heat capacity}\times\text{temperature difference}\\\\E_{P}=m\times{C_{W}}\times\delta{T}$....equation-1

Mass of the water (m) = 100 cc/sec = 100 g/sec

Specific heat capacity of water =$C_{w}=4.2$j/gm

Temperature difference $\delta{T}=(T_{2}-T_{1})$ Where $T_{2}$ is the final temperature and $T_{1}$ is the initial temperature of water.

Therefore,

$E_{P}=m\times{C_{w}}\times\delta{T}\\\\\Rightarrow1600=100\times4.2\times({T_{2}-10)$

$\Rightarrow{T_{2}-10=\frac{1600}{4.2\times100}$

$\Rightarrow{T_{2}}=\frac{1600}{420}+10=3.80+10=13.80$

The outlet temperature will be 13.8 °C