Question

A cube of water 10 cm on a side is placed in a microwave beam having $E_0=11kV/m$. The microwaves illuminate one face of the cube, and the water absorbs $80\%$of the incident energy. How long will it take to raise the water temperature by $50˚C?$.Assume that the water has no heat loss during this time.

Short answer

The raise of the water temperature is$\Delta t=162\mathrm{s}$

Step-by-step solution

Definition of Electromagnetic fields and Waves:

A charge near its surroundings produces an electric field, which is a force. When other charges are brought near this field, they are subjected to this force. N/C (Force/Charge) is the SI unit for electric fields.

Given values:

$$\begin{gathered} A=0.10m \\ {E}_0=11kV/m \\ T=50˚ \end{gathered}$$

Substitute the rate of energy values into equation:

$P=(80\%)IA$

$P=(0.80)\frac{c{\epsilon }_o}{2}{E}_o^{2}A$

$P=(0.80)\frac{\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2\right)}{2}{\left(11\times {10}^3\mathrm{V}/\mathrm{m}\right)}^2(0.10\mathrm{m})$

localid="1649933986875" $P=1.29kW$

Equation:

Now, Find value for $△E$:

$\Delta E=mc\Delta T$

$\Delta E=(0.10\mathrm{m}{)}^3\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\left(4186\mathrm{J}/\mathrm{kg}/{}^{\circ }\mathrm{C}\right)\left(50°\mathrm{C}\right)$

$\Delta E=2.10\times {10}^5\mathrm{J}$

Substitute the values in equation: 

Finally, find the values for $△t$:

$\Delta t=\frac{\Delta E}{P}$

$$\begin{gathered} \Delta t=\frac{2.10\times {10}^5\mathrm{J}}{1.29\times {10}^{-3}\mathrm{W}} \\ △t=162s \end{gathered}$$

Final Result:

The change in the quantity of water is$△t=162s$.