
An electric heater of resistance 8\[\Omega \] draws 15A from the service mains in 2hrs. Calculate the rate at which heat is developed in the heater.
Answer
504.9k+ views
Hint: We are given the information of resistance, current and the time taken by the device to operate with the given parameters. We can easily find the heat developed in the heater during this time using the Joule’s law of heating in electricity.
Complete answer:
We know that any device which uses the electrical power to be converted to mechanical energy produces a good deal of heat energy. The Joule’s law of heating comes with this principle. It states that the heat energy developed in an operating device is proportional to the resistance, square of the current and the time taken. The Joule’s law of heat is expressed mathematically as –
\[{{E}_{heat}}={{I}^{2}}Rt\]
We can express this even in the terms of voltage or power, relating them with the current flowing through the device. The Joule’s heating involved in terms of voltage is given by –
\[\begin{align}
& V=IR \\
& \Rightarrow \text{ }{{\text{E}}_{heat}}=\dfrac{{{V}^{2}}}{R}t \\
\end{align}\]
Also, we can express it in terms of the power as –
\[\begin{align}
& P=VI={{I}^{2}}R \\
& \Rightarrow {{E}_{heat}}=Pt \\
\end{align}\]
Now, let us consider the given situation. The resistance of the heater is given as –
\[R=8\Omega \]
The current drawn from the mains source is given as 15A.
The time duration of operation is given as 2hrs.
So, the rate at which the heat developed by Joule’s heating is the power rating of the device, which is given by –
\[\begin{align}
& {{P}_{heat}}={{I}^{2}}R \\
& \Rightarrow \text{ }{{\text{P}}_{heat}}={{(15A)}^{2}}.(8\Omega ) \\
& \Rightarrow \text{ }{{\text{P}}_{heat}}=1800W \\
\end{align}\]
The rate at which the heat is developed is given as 1.8kW.
Note:
We can directly substitute for the power to find the rate of heat developed in the device. The above descriptive method helps us understand that the rate of the heat developed is nothing other than the power rating of the device from a constant main source.
Complete answer:
We know that any device which uses the electrical power to be converted to mechanical energy produces a good deal of heat energy. The Joule’s law of heating comes with this principle. It states that the heat energy developed in an operating device is proportional to the resistance, square of the current and the time taken. The Joule’s law of heat is expressed mathematically as –
\[{{E}_{heat}}={{I}^{2}}Rt\]
We can express this even in the terms of voltage or power, relating them with the current flowing through the device. The Joule’s heating involved in terms of voltage is given by –
\[\begin{align}
& V=IR \\
& \Rightarrow \text{ }{{\text{E}}_{heat}}=\dfrac{{{V}^{2}}}{R}t \\
\end{align}\]
Also, we can express it in terms of the power as –
\[\begin{align}
& P=VI={{I}^{2}}R \\
& \Rightarrow {{E}_{heat}}=Pt \\
\end{align}\]
Now, let us consider the given situation. The resistance of the heater is given as –
\[R=8\Omega \]
The current drawn from the mains source is given as 15A.
The time duration of operation is given as 2hrs.
So, the rate at which the heat developed by Joule’s heating is the power rating of the device, which is given by –
\[\begin{align}
& {{P}_{heat}}={{I}^{2}}R \\
& \Rightarrow \text{ }{{\text{P}}_{heat}}={{(15A)}^{2}}.(8\Omega ) \\
& \Rightarrow \text{ }{{\text{P}}_{heat}}=1800W \\
\end{align}\]
The rate at which the heat is developed is given as 1.8kW.
Note:
We can directly substitute for the power to find the rate of heat developed in the device. The above descriptive method helps us understand that the rate of the heat developed is nothing other than the power rating of the device from a constant main source.
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