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
Verified
118.2k+ views
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.
$\eta =\dfrac{P}{{{P}_{r}}}$
Given, efficiency $\eta =80%=0.8$
Actual power, $P=\eta {{P}_{r}}=0.8\times 2=1.6kW$
This power is used to heat the water with the flow-rate $Q = 100 cc/sec.$
The relationship between mass and volume is given by –
$\rho =\dfrac{m}{v}$
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, ${m}'=100\dfrac{g}{cc}=0.1\dfrac{kg}{cc}$
The corresponding rate of heat transferred to the flow of water which raises its temperature by $\Delta T$ is given by –
Rate of heat transfer, ${Q}'={m}'c\Delta T$
where c = specific heat capacity of water = $4180 J/kg ^{\circ }C$
The rate of heat transfer is equal to the actual power produced at the geyser.
${Q}'=P$
$\Rightarrow P={m}'c\Delta T$
Substituting, we get –
$1.6\times {{10}^{3}}=0.1\times 4180\times \Delta T$
Solving for $\Delta T$, we get –
$\Delta T=\dfrac{1.6\times {{10}^{3}}}{0.1\times 4180}=3.82$
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.
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.
$\eta =\dfrac{P}{{{P}_{r}}}$
Given, efficiency $\eta =80%=0.8$
Actual power, $P=\eta {{P}_{r}}=0.8\times 2=1.6kW$
This power is used to heat the water with the flow-rate $Q = 100 cc/sec.$
The relationship between mass and volume is given by –
$\rho =\dfrac{m}{v}$
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, ${m}'=100\dfrac{g}{cc}=0.1\dfrac{kg}{cc}$
The corresponding rate of heat transferred to the flow of water which raises its temperature by $\Delta T$ is given by –
Rate of heat transfer, ${Q}'={m}'c\Delta T$
where c = specific heat capacity of water = $4180 J/kg ^{\circ }C$
The rate of heat transfer is equal to the actual power produced at the geyser.
${Q}'=P$
$\Rightarrow P={m}'c\Delta T$
Substituting, we get –
$1.6\times {{10}^{3}}=0.1\times 4180\times \Delta T$
Solving for $\Delta T$, we get –
$\Delta T=\dfrac{1.6\times {{10}^{3}}}{0.1\times 4180}=3.82$
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.
Recently Updated Pages
A team played 40 games in a season and won 24 of them class 9 maths JEE_Main
Here are the shadows of 3 D objects when seen under class 9 maths JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
Madhuri went to a supermarket The price changes are class 9 maths JEE_Main
If ax by czand b2 ac then the value of yis 1dfrac2xzleft class 9 maths JEE_Main
Trending doubts
JEE Main 2025: Application Form (Out), Exam Dates (Released), Eligibility & More
JEE Main Login 2045: Step-by-Step Instructions and Details
Class 11 JEE Main Physics Mock Test 2025
JEE Main Chemistry Question Paper with Answer Keys and Solutions
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
Other Pages
NCERT Solutions for Class 11 Physics Chapter 7 Gravitation
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids
Units and Measurements Class 11 Notes - CBSE Physics Chapter 1
NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs