Question

hii ,,,

a continuous flow water heater geyser has an electrical power rating 2 kilowatt and efficiency of conversion of electric power into heat 80% if water is flowing through the device @ of 100 CC per second and the inlet temperature is 10 degree centigrade then the 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 =$C_{W}=4.2\text{J/gm}$

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

Answer 2

Given:

Electrical power is 2 KW which is 2000 W and the efficiency is of 80 percent therefore the heat absorbed by the water is $E = 2000\times \frac {80}{100} = 1600 J$

As we know that the heat energy is calculated by the following formula,

$Heat \quad energy = Mass \quad \times \quad Specific \quad heat \quad capacity \quad \times \quad Temperature \quad Difference$

$\Rightarrow E = m \times C w \times\partial T$

Given data:

Mass of the water be 100 cc per sec which is 100 g per sec.

Cw is 4.2 J/gm.

$\partial T=T2-T1$

Thereby, $1600 = 100 \times 4.2\times (T2 - 10)$

$\Rightarrow T2 -10 = \frac {1600 }{4.2 }\times 100$

$\Rightarrow T2 = 3.80 + 10$

T2 = 13.8 degree Celsius