(a) [tex]1.63\cdot 10^5 mol[/tex]
First of all, we need to calculate the energy needed for to raise the temperature of the water:
[tex]Q=m C_S \Delta T[/tex]
where
m = 345 g is the mass of the water
Cs = 4.186 g/J C is the specific heat of water
[tex]\Delta T=99.8C-26.5C=73.3^{\circ}C[/tex] is the change in temperature of the water
Substituting,
[tex]Q=(345 g)(4.186 J/gC)(73.3 C)=1.06\cdot 10^5 J[/tex]
Then we can calculate the energy of one microwave photon with wavelength
[tex]\lambda=12.2 cm=0.122 m[/tex]:
[tex]E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{0.122 m}=1.63\cdot 10^{-24} J[/tex]
where h is the Planck constant and c is the speed of light
So the total number of photons needed is
[tex]n=\frac{Q}{E}=\frac{1.06\cdot 10^5 J}{1.63\cdot 10^{-24} J}=9.8\cdot 10^{28}J[/tex]
And the corresponding number of moles is
[tex]n=\frac{N}{N_A}=\frac{9.8 \cdot 10^{28}J}{6.022\cdot 10^{23} }=1.63\cdot 10^5 mol[/tex]
where [tex]N_A[/tex] is the Avogadro number (the number of photons in 1 mole).
(b) 151.4 s
The power delivered by the microwave oven is
P = 700 W
We know that the power is the ratio of the energy delivered by the time taken:
[tex]P=\frac{E}{t}[/tex]
since in this case the total energy needed to heat the water is [tex]1.06\cdot 10^5 J[/tex], then the time needed is
[tex]t=\frac{E}{P}=\frac{1.06\cdot 10^5 J}{700 W}=151.4 s[/tex]
