
An electric immersion heater of $1.08kW$ is immersed in water. After the water has reached a temperature of ${{100}^{0}}C$ , how much time will be required to produce $100g$ steam?
Answer
537.3k+ views
Hint: We have been given the power supplied to the water by the immersion rod heats the water up to ${{100}^{0}}C$ . Now, on further heating the water, the temperature of water will not increase but its state will change from water to water vapour. This energy that is being absorbed by the water to change its state is called Latent heat of Vaporization of water and its value at $1atm$ and ${{100}^{0}}C$ is $2.26kJ{{g}^{-1}}$
Complete step by step answer:
We know that, the Latent heat of vaporization of water is equal to:
$=2.26kJ{{g}^{-1}}$
Which can also be written as:
$=2260J{{g}^{-1}}$
Now, this is the amount of heat energy required to convert 1g of water into 1g of water vapour. Therefore, energy required to convert 100g of water into water vapour is equal to:
$\begin{align}
& =2260\times 100J \\
& =226000J ---(1)\\
\end{align}$
Now, the power output of the immersion rod is given to us as:
$=1.08kW$
Which means the energy output of the immersion rod is:
$\begin{align}
& =1.08\times {{10}^{3}}J{{s}^{-1}} \\
& =1080J{{s}^{-1}} ---(2)\\
\end{align}$
Therefore, time taken (say t) to convert 100g water into 100g steam can be found out by dividing equation number (1) by equation number (2). Thus we get:
$\begin{align}
& \Rightarrow t=\dfrac{226000}{1080}s \\
& \Rightarrow t=209.26s \\
\end{align}$
Therefore, the amount taken by the immersion rod to convert 100g water into steam is $209.26g$.
Note: The value of terms like specific heat of water, specific heat of ice or steam and latent heat of water at different state conversion should be known to us. It should also be noted that, when a certain amount of any substance is undergoing a change in its state, all the energy absorbed or released by the substance is used in changing its state of matter first and only after all of it has entered a new state, the energy will then be used to increase or decrease its temperature.
Complete step by step answer:
We know that, the Latent heat of vaporization of water is equal to:
$=2.26kJ{{g}^{-1}}$
Which can also be written as:
$=2260J{{g}^{-1}}$
Now, this is the amount of heat energy required to convert 1g of water into 1g of water vapour. Therefore, energy required to convert 100g of water into water vapour is equal to:
$\begin{align}
& =2260\times 100J \\
& =226000J ---(1)\\
\end{align}$
Now, the power output of the immersion rod is given to us as:
$=1.08kW$
Which means the energy output of the immersion rod is:
$\begin{align}
& =1.08\times {{10}^{3}}J{{s}^{-1}} \\
& =1080J{{s}^{-1}} ---(2)\\
\end{align}$
Therefore, time taken (say t) to convert 100g water into 100g steam can be found out by dividing equation number (1) by equation number (2). Thus we get:
$\begin{align}
& \Rightarrow t=\dfrac{226000}{1080}s \\
& \Rightarrow t=209.26s \\
\end{align}$
Therefore, the amount taken by the immersion rod to convert 100g water into steam is $209.26g$.
Note: The value of terms like specific heat of water, specific heat of ice or steam and latent heat of water at different state conversion should be known to us. It should also be noted that, when a certain amount of any substance is undergoing a change in its state, all the energy absorbed or released by the substance is used in changing its state of matter first and only after all of it has entered a new state, the energy will then be used to increase or decrease its temperature.
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