Question

A continuous flow water heater (geyser) has an electrical power rating equals \[2\,{\text{kW}}\] and efficiency of electric power into heat equals \[80\% \] . If water is flowing through the device at the rate of \[100\,{\text{cc}}\,{{\text{s}}^{ - 1}}\] , the inlet temperature is \[10\,^\circ {\text{C}}\] , the outlet temperature will be:
A. \[12.2\,^\circ {\text{C}}\]
B. \[13.8\,^\circ {\text{C}}\]
C. \[20\,^\circ {\text{C}}\]
D. \[16.5\,^\circ {\text{C}}\]

Answer 1

Hint: First of all, we will find the effective power from the percentage given. After that we will use the formula of heat which relates specific heat, mass and change in temperature. Heat can be obtained from power itself. We will manipulate accordingly to obtain the result.

Formula used:
The formula which gives us the heat is given below:
\[H = m \times c \times \Delta T\] …… (1)
Where,
\[H\] indicates the heat generated.
\[m\] indicates the mass of the water.
\[c\] indicates the specific heat capacity.
\[\Delta T\] indicates the difference in temperature.

Complete step by step answer:
In the given question, we are supplied the following data:
The rated electric power of a geyser is \[2\,{\text{kW}}\] .The efficiency of the conversion of electric power is \[80\% \] .The volume of water flowing per unit time is \[100\,{\text{cc}}\,{{\text{s}}^{ - 1}}\] . The inlet temperature is \[10\,^\circ {\text{C}}\] .We are asked to find the temperature of the outlet.

To begin with, we will try to find the mass of the water.
As we know, the density of water is \[1\,{\text{g}}\,{\text{c}}{{\text{m}}^{ - 3}}\] .Since, the volume of water flowing per unit time is \[100\,{\text{cc}}\,{{\text{s}}^{ - 1}}\] , so the mass of water is \[100\,{\text{g}}\] .

The efficient power is given by: (with \[80\% \] efficiency)
$P = \dfrac{{80}}{{100}} \times 2\,{\text{kW}} \\
\Rightarrow P = 1.6\,{\text{kW}} \\
\Rightarrow P = 1600\,{\text{W}} \\$
Now, we can say that work done \[1600\,{\text{J}}\] is done in one second.
We will now, substitute the required values in the equation (1) and we get:
$H = m \times c \times \Delta T \\
\Rightarrow \Delta T = \dfrac{H}{{m \times c}} \\
\Rightarrow \Delta T = \dfrac{{1600}}{{100 \times 4.186}} \\
\therefore \Delta T = 3.8\,^\circ {\text{C}} \\$
Therefore, the difference in temperature is found to be \[3.8\,^\circ {\text{C}}\] .
So, the temperature of the outlet is:
$10\,^\circ {\text{C}} + 3.8\,^\circ {\text{C}} \\
\therefore 13.8\,^\circ {\text{C}}$

The correct option is B.

Note: In the given question, most of the students seem to have confusion regarding the amount of heat energy involved in the process. It is important to remember that the work done and the heat energy are both equivalent quantities. Power is nothing but the work done per unit time.