Question

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

Answer 1

Hint: In Physics, energy can be converted from one form to another, but the quantity energy remains the same. Here, we can obtain the answer by equating the quantity of electrical energy and heat energy since the quantity is the same but the form is different.

Complete step by step answer:
The electric geyser is an electrical appliance used to raise the temperature of water using electric current. It converts electrical energy to thermal energy.
The power rating of the geyser is given, along with the efficiency. The power rating determines the amount of energy converted to heat per second.
Power rating, $P_r=2kW$
The coil used by the geyser to heat the water experiences some electric losses. Hence, the actual power obtained by the geyser is lesser than the rated power.
Efficiency is defined as the ratio of actual power to the rated power.
$\eta ={\displaystyle \frac{P}{P_r}}$
Given, efficiency $\eta =80$
Actual power, $P=\eta P_r=0.8\times 2=1.6kW$
This power is used to heat the water with the flow-rate $Q=100cc/sec.$
The relationship between mass and volume is given by –
$\rho ={\displaystyle \frac{m}{v}}$
where $\rho$= density.
Density of water at room temperature is 1 gram/cc. Hence, the mass occupied by 100cc volume of water = 100 g.
Hence, mass flow rate, $m^{\prime }=100{\displaystyle \frac{g}{cc}}=0.1{\displaystyle \frac{kg}{cc}}$
The corresponding rate of heat transferred to the flow of water which raises its temperature by $\mathrm{\Delta }T$ is given by –
Rate of heat transfer, $Q^{\prime }=m^{\prime }c\mathrm{\Delta }T$
where c = specific heat capacity of water = $4180J/k{g}^{\circ }C$
The rate of heat transfer is equal to the actual power produced at the geyser.
$Q^{\prime }=P$
$\Rightarrow P=m^{\prime }c\mathrm{\Delta }T$
Substituting, we get –
$1.6\times {10}^3=0.1\times 4180\times \mathrm{\Delta }T$
Solving for $\mathrm{\Delta }T$, we get –
$\mathrm{\Delta }T={\displaystyle \frac{1.6\times {10}^3}{0.1\times 4180}}=3.82$
Given that the inlet temperature is ${10}^{\circ }C$ and change in temperature is $3.82$,
The outlet temperature = $13.8^{\circ }C$

Hence, the correct option is Option B.

Note: The SI unit of specific heat capacity is J/kg/${}^{\circ }C$. The other commercial units of specific heat capacity are called small calorie and big calorie. Small calorie, also called cal, is equal to 4.18 J and big calorie is the kilocalorie, represented by kCal, which is equal to 4180 joules.