A cube of water 10 cm on a side is placed in a microwave beam having $E_{0}=11 \mathrm{kV} / \mathrm{m}$. The microwaves illuminate one face of the cub e, a nd the water absorbs 80% of the incident energy. How long will it take to raise the water temperature by $50^{\circ} \mathrm{C}$? Assume that the water has no heat loss during this time.

$\textbf{Given values:}$

$A=0.10 \: \text{m}$

$E_{o}=11 \: \text{kV}/\text{m}$

$T=50 ^\circ$

The rate of energy transfer :

$$\begin{align*} P&=(80 \%) I A \\ P&=(0.80) \frac{c \varepsilon_{o}}{2} E_{o}^2 A\\ P&=(0.80) \frac{(3 \times 10^8 \: \text{m}/\text{s})(8.85 \times 10^{-12} \: \text{C}^2/\text{N} \cdot \text{m}^2)}{2}(11 \times 10^3 \: \text{V}/\text{m})^2(0.10 \: \text{m}) \tag{Substitute values in equation.}\\ P&=1.29 \: \text{kW}\\ \end{align*}$$

Now, we find value for $\Delta E$ :

$$\begin{align*} \Delta E&=mc \Delta T \\ \Delta E&=(0.10 \: \text{m})^3 (1000 \: \text{kg}/\text{m}^3)(4186 \: \text{J}/\text{kg} / ^\circ \text{C})(50 ^\circ \text{C}) \tag{Substitute values in equation.}\\ \Delta E&=2.10 \times 10^5 \: \text{J}\\ \end{align*}$$

Finally, we find value for $\Delta t$ :

$$\begin{align*} \Delta t&=\frac{\Delta E}{P}\\ \Delta t&=\frac{2.10 \times 10^5 \: \text{J}}{1.29 \times 10^{-3} \: \text{W}} \tag{Substitute values in equation.}\\ \Delta t&=162 \: \text{s}\\ \end{align*}$$