Question

# Specific heat of water is $4.2J{{(g{}^\circ C)}^{-1}}$. If light of frequency $3\times {{10}^{9}}Hz$ is used to heat $400g$ of water from $20{}^\circ C$to $40{}^\circ C$, the number of photons needed will be?

Hint: The energy required by water to increase its temperature can directly be calculated using the formula of specific heat of water. At the same time, energy of photons is calculated. The number of photons required to heat the water is equal to the ratio of energy required to increase the temperature to the energy of photons.

Formula used:
\begin{align} & 1)Q=m{{C}_{p}}\Delta T \\ & 2){{E}_{p}}=h\nu \\ & 3)n=\dfrac{Q}{{{E}_{p}}} \\ \end{align}
Here,
$Q$ is the energy required by the water to increase its temperature by $T{}^\circ C$
$m$ is the mass of water
${{C}_{p}}$ is the specific heat of water
$\Delta T$ is the change in temperature
${{E}_{p}}$ is the energy of photons
$h$ is the Planck’s constant
$\nu$ is the frequency of light
$n$ is the number of photons

Specific heat of water refers to the amount of heat required to increase the temperature of $1g$ of water by $1{}^\circ C$. Specific heat formula is given by
$Q=m{{C}_{p}}\Delta T$
Here, $Q$ is the energy required to increase the temperature of $m$ grams of water by $\Delta T{}^\circ C$.$\Delta$ refers to the change in temperature.

We have

$m=400g$
${{C}_{p}}=4.2J{{(g{}^\circ C)}^{-1}}$
$\Delta T={{T}_{2}}-{{T}_{1}}=20{}^\circ C$ $\left[ {{T}_{2}}=40{}^\circ C,{{T}_{1}}=20{}^\circ C \right]$

On substituting these values,

$Q=m{{C}_{p}}\Delta T=400\times 4.2\times 20=33.6\times {{10}^{3}}J$
Now, let us move on to the next formula. The energy of photons is given by
${{E}_{p}}=h\nu$

We have

$h=6.62\times {{10}^{-34}}$
$\nu =3\times {{10}^{9}}Hz$

Substituting these values in the above equation,

${{E}_{p}}=h\nu =6.62\times {{10}^{-34}}\times 3\times {{10}^{9}}=19.86\times {{10}^{-25}}J$
Finally, the number of photons required to heat water is given by
$n=\dfrac{Q}{{{E}_{p}}}=\dfrac{33.6\times {{10}^{3}}}{19.86\times {{10}^{-25}}}=1.69\times {{10}^{28}}$

So, the correct answer is “Option B”.

Note:
Students should take care of the units of parameters mentioned in the question. Sometimes, conversions are required to calculate the answer. In this question, conversions were not required since all the units were matching and cancelled out each other most of the time. It is also noted that direct formulas are asked to determine simple additional parameters. So, it is important to remember direct formulas along with simple additional formulas to answer such types of questions.