Answer
$1.2\times 10^{19}~photons$
Work Step by Step
We can find the required number of photons as follows:
$E=mc_{water}\Delta T$
We plug in the known values to obtain:
$E=(1.0\times 10^{-3}Kg)(4190J/Kg.C^{\circ})(1.0C^{\circ})$
$E=4.19J$
Now $n=\frac{E\lambda}{hc}$
We plug in the known values to obtain:
$n=\frac{(4.19J)(550\times 10^{-9}m)}{(6.63\times 10^{-34}J.s)(3.0\times 10^8m/s)}$
$n=1.2\times 10^{19}~photons$
