Question

A cube of water 10 cm on a side is placed in a microwave beam having Ea = 11 kV/m‘ The microwaves illuminate one faceofthe cube, and the water absorbs 80% of the incident energy, How long will it take to raise the water temperature bySO'C? Assume that the water has no heal loss during this time.

Answer 1

Answer:

The time is 133.5 sec.

Explanation:

Given that,

One side of cube = 10 cm

Intensity of electric field = 11 kV/m

Suppose How long will it take to raise the water temperature by 41°C Assume that the water has no heat loss during this time.

We need to calculate the rate of energy transfer from the beam to the cube

Using formula of rate of energy

$P=(0.80)IA$

$P=0.80\times\dfrac{c\mu_{0}E^2}{2}\times A$

Put the value into the formula

$P=0.80\times\dfrac{3\times10^{8}\times8.85\times10^{-12}\times(1.1\times10^{4})^2}{2}\times(10\times10^{-2})^2$

$P=1285.02\ W$

We need to calculate the amount of heat

Using formula of heat

$E =mc\Delta T$

$E =\rho Vc\Delta T$

Put the value into the formula

$E=1000\times(0.10)^3\times4186\times41$

$E=171626\ J$

We need to calculate the time

Using formula of time

$t=\dfrac{E}{P}$

Put the value into the formula

$t=\dfrac{171626}{1285.02}$

$t=133.5\ sec$

Hence, The time is 133.5 sec.