Question

A cube of water 10 cm on a side is placed in a microwave beam having E0=11kV/mE_{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 50C50^{\circ} \mathrm{C}? Assume that the water has no heat loss during this time.

Solution

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Given values:\textbf{Given values:}

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

Eo=11kV/mE_{o}=11 \: \text{kV}/\text{m}

T=50T=50 ^\circ

The rate of energy transfer :

P=(80%)IAP=(0.80)cεo2Eo2AP=(0.80)(3×108m/s)(8.85×1012C2/Nm2)2(11×103V/m)2(0.10m)P=1.29kW\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 ΔE\Delta E :

ΔE=mcΔTΔE=(0.10m)3(1000kg/m3)(4186J/kg/C)(50C)ΔE=2.10×105J\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 Δt\Delta t :

Δt=ΔEPΔt=2.10×105J1.29×103WΔt=162s\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*}

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