At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
Given,
The current drawn by the heating element, I=15 A
The supply voltage, V=220 V
The mass of the water, m=3 kg
The temperature of the water, T=100 °C
The latent heat of vaporization of water, L_v=2.25×10⁶ J/kg
The power of the heating element is given by,
[tex]P=IV=\frac{W}{t}[/tex]Where W is the work done by the heating element on the water and t is the time it takes for the heating element to evaporate the water.
On rearranging,
[tex]t=\frac{W}{IV}[/tex]The work done by the heater is the same as the work done on the water. And the work done on the water is equal to the required to evaporate the given mass of water.
i.e.,
[tex]W=mL_v[/tex]Therefore,
[tex]t=\frac{mL_v}{IV}[/tex]As all the quantities are in SI units the time will be in seconds.
Thus, the time it takes to evaporate the water in minutes is given by,
[tex]t=\frac{mL_v}{IV\times60}[/tex]On substituting the known values,
[tex]\begin{gathered} t=\frac{3\times2.25\times10^6}{15\times220\times60} \\ =34.1\text{ min} \end{gathered}[/tex]Therefore the time it takes to evaporate the given amount of water is 34.1 min
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
Six people to can perform a task in 8 days. If we add 2 equally able people, how long would it take?
