Question

A cup of vol. 0.20, radius= 0.30m is filled with room temperature water at 21 C. Yiu want to boill the water in microwave oven in 3.00 min. Assume all the microwaves come uniformly from the top of the oven and are completely absorbed by the water ( specific heat for Water is c=4181 j/(k kg) and the dentisty of water is 1000 kg/m3) the required are 1- Average power? 2- Average intensity? 3- rm( root mean square) electric field of the microwaves?

Answer 1

1).Volume of cup= 0.20m³

Radius of cup,r=0.30m

Room temerature, T=21C= (21+273)= 294 K

Density of water, $\rho= 1000kg/m^3$

Total mass of the water in the given cup, $m=V\rho= 0.20 \times 1000= 200kg$ ( THIS MUCH MASS IS NOT PRACTICAL. SO PLEASE CONFIRM THE UNITS YOU PROVIDED IN THE QUESTION)

Specific heat of water,

Final temperature of the water in boiling is, T_f= 100 degrees celsius= (100+273)K= 373K

Heat enerfy needed by water for boiling is, $Q=mC(T_f-Ti)=200 \times 4181 \times (373-294)= 200 \times 4181 \times 79$

$Q=6.605 \times 10^7 J$

Microwave must provide this much heat energy to have boiling.

Now Average power,P= Energy required/ Time

   $P=Q/t= (6.605 \times 10^7)/(3 \times 60)= 3.6666 \times 10^5J/s$

   $P= 366.66 \times 10^3 watt= 366.66 KW$

2). Now, Average intensity(I_avg)= Power/ Area= P/A

Given: r= 0.30m

then, Area $A=\pi r^2= 3.14\times (0.30)^2= 0.2826 m^2$

thus,    $I_{avg}=P/A= (366.66 \times 10^3)/ 0.2826 = 1.297 \times 10^6 W/m^2$

3). Also, $I_{avg}=c\epsilon_{0}E_{0}^2/2$ .................(1)

   where c= speed of light = 3* 10⁸ m/s

   $\epsilon_{0}= 8.85 \times 10^{-12} C^2N^{-1}m^{-2}$

   E₀ = peak electric strength and

   $E_{rms}= E_{0}/\sqrt{2}$

then using equation 1

   $E_{0}=\sqrt{(2I_{avg})/(c\epsilon_{0})}$

or, $E_{rms}=(\sqrt{(2I_{avg})/(c\epsilon_{0})})/\sqrt{2}$

   $E_{rms}=\sqrt{(2I_{avg})/(2c\epsilon_{0})}$

   $E_{rms}=\sqrt{(I_{avg})/(c\epsilon_{0})}$

   $E_{rms}=\sqrt{(1.297 \times 10^{6})/(3 \times 10^8 \times 8.85 \times 10^{-12})}$

   $E_{rms}=\sqrt{(1.297 \times 10^{6})/(26.55 \times 10^{-4})}= \sqrt{4.885 \times 10^8}$

      $E_{rms}=2.21020 \times 10^4 V/m$ ( ANS)