Respuesta :
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
