Question

Specific heat of water is $4.2\text{J/kg K}$. If light of frequency $4\times {10}^9\text{Hz}$ is used to heat $100\text{gm}$ of water from $20{}^{\circ }\text{C}$ to $100{}^{\circ }\text{C}$. The no. of photons required will be,

• A. $1.27\times {10}^{28}$
• B. $1.6\times {10}^{24}$
• C. $6.36\times {10}^{14}$
• D. $2.80\times {10}^5$

Answer 1

The correct option is A $1.27\times {10}^{28}$

Heat required is, $H=ms\Delta T$
or, $H=100\times 4.2\times (100-20)=33.6\times {10}^3\text{J}$
Energy of each photon will be,
$E=hf$
or, $E=6.63\times {10}^{-34}\times 4\times {10}^9=2.65\times {10}^{-24}\text{J}$

Therefore, no. of photons required to heat the water is given as
$n=\frac{H}{\text{Energy of one photon}}=\frac{33.6\times {10}^3}{2.65\times {10}^{-24}}=1.27\times {10}^{28}$

Hence, $\left(A\right)$ is the correct answer.

$$\begin{array}{c}\text{Why this Question ?}\\ \text{This is a combined question from thermal and photoelectron emission. Concept: energy=}mc\Delta T\text{=ntimes hf}\end{array}$$