# Question 2e81a

Apr 12, 2016

$2.17 \cdot {10}^{28} \text{photons}$

#### Explanation:

The first thing to do here is determine how much energy in the form of heat is required to increase the temperature of that sample of coffee from ${25.0}^{\circ} \text{C}$ to ${62.0}^{\circ} \text{C}$.

Your tool of choice here will be the following equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} q = m \cdot c \cdot \Delta T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$q$ - the amount of heat gained / lost
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature, defined as the difference between the final temperature and the initial temperature

Notice that the problem provides you with the volume of the sample. Use the coffee's density to find the mass of this sample

225 color(red)(cancel(color(black)("mL"))) * overbrace("0.997 g"/(1color(red)(cancel(color(black)("mL")))))^(color(purple)("given density of coffee")) = "224.325 g"

In you case, the change in temperature will be equal to

$\Delta T = {62.0}^{\circ} \text{C" - 25.0^@"C" = 37.0^@"C}$

The specific heat of coffee is given to you in ${\text{J g"^(-1)"K}}^{- 1}$. It's worth noting that this will be equivalent to ${\text{J g"^(-1)""^@"C}}^{- 1}$, so you don't need to convert the temperature from degrees Celsius to Kelvin.

Plug in your values in the above equation to find the heat needed to heat the coffee

$q = 224.325 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 4.184"J"/(color(red)(cancel(color(black)("g")))color(red)(cancel(color(black)(""^@"C")))) * 37color(red)(cancel(color(black)(""^@"C}}}}$

$q = \text{34,727.3 J}$

Your goal now will be to find the energy of a single photon of wavelength $\text{12.4 cm}$. To do that, use the Planck - Einstein relation, which states that the energy of a photon is proportional to its frequency

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} E = h \cdot \nu \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

$E$ - the energy of the photon
$\nu$ - its frequency
$h$ - Planck's constant, equal to $6.626 \cdot {10}^{- 34} \text{J s}$

Now, frequency and wavelength have an inverse relationship described by the equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \nu \cdot l a m \mathrm{da} = c \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

$l a m \mathrm{da}$ - the wavelength of the photon
$c$ - the speed of light in a vacuum, usually given as $3 \cdot {10}^{8} {\text{m s}}^{- 1}$

Use this equation to find the frequency of a photon that has a wavelength of $\text{12.4 cm}$. Notice that the speed of light is expressed in meters per second, so make sure that you convert the wavelength from centimeters to meters

$l a m \mathrm{da} \cdot \nu = c \implies \nu = \frac{c}{l a m \mathrm{da}}$

Plug in your value to get

nu = (3 * 10^8 color(red)(cancel(color(black)("m")))"s"^(-1))/(12.4 * 10^(-2)color(red)(cancel(color(black)("m")))) = 2.419 * 10^9"s"^(-1)

This means that the energy of a single photon of microwave radiation will be

E = 6.626 * 10^(-34)"J" color(red)(cancel(color(black)("s"))) * 2.419 * 10^9color(red)(cancel(color(black)("s"^(-1))))

$E = 1.603 \cdot {10}^{- 24} \text{J}$

Now all you have to do is figure out how many photons of microwave radiation are needed to get a total energy of $\text{34,727.3 J}$

"34,727.3" color(red)(cancel(color(black)("J"))) * "1 photon"/(1.603 * 10^(-24)color(red)(cancel(color(black)("J")))) = color(green)(|bar(ul(color(white)(a/a)2.17 * 10^(28)"photons"color(white)(a/a)|)))#

The answer is rounded to three sig figs.