Question

A certain microwave oven delivers $750.$ watts $(\mathrm{J}/\mathrm{s})$ of power to a coffee cup containing $50.0\mathrm{g}$ water at ${25.0}^{\circ }\mathrm{C}$. If the wavelength of microwaves in the oven is $9.75\mathrm{cm}$, how long does it take, and how many photons must be absorbed, to make the water boil? The specific heat capacity of water is $4.18\mathrm{J}/{}^{\circ }\mathrm{C}\cdot \mathrm{g}$ and assume only the water absorbs the energy of the microwaves.

Short answer

It takes approximately 20.93 seconds to heat the water to boiling point, and during this process, approximately $7.70\times {10}^{27}$ photons need to be absorbed.

Step-by-step solution

Calculate the energy required to heat the water to boiling point

First, we need to find the energy required to raise the temperature of the water from the initial temperature (25°C) to the boiling point (100°C). To do this, we can use the specific heat capacity formula: $Q=mc\Delta T$ where: $Q$ = heat energy (Joules) $m$ = mass of water (grams) $c$ = specific heat capacity of water ($4.18\mathrm{J}/{}^{\circ }\mathrm{C}\cdot \mathrm{g}$) $\Delta T$ = change in temperature ($100-25$) Plugging in the values, we get: $Q=(50.0)(4.18)(100-25)$ $Q=15700\mathrm{J}$ The energy required to heat the water to boiling point is 15,700 Joules.

Calculate the time it takes to deliver the required energy

Since we are given the power of the microwave oven, which is the rate of energy delivery, we can find the time it takes to deliver the required energy using the following formula: $t=\frac{Q}{P}$ where: $t$ = time (seconds) $Q$ = heat energy (Joules) $P$ = power of the microwave (Watts, which is Joules per second) Plugging in the values, we get: $t=\frac{15700}{750}$ $t=20.93\mathrm{s}$ It takes approximately 20.93 seconds to heat the water to boiling point.

Calculate the energy per photon

Next, we calculate the energy per photon using the formula: $E=\frac{hc}{\lambda }$ where: $E$ = energy per photon (Joules) $h$ = Planck's constant ($6.626\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}$) $c$ = speed of light ($3.00\times {10}^8\mathrm{m}/\mathrm{s}$) $\lambda$ = wavelength of the microwaves (in meters) First, convert the wavelength to meters by dividing by 100: $\lambda =\frac{9.75}{100}=0.0975\mathrm{m}$ Plugging in the values, we get: $E=\frac{(6.626\times {10}^{-34})(3.00\times {10}^8)}{0.0975}$ $E=2.040\times {10}^{-24}\mathrm{J}$ The energy per photon is approximately $2.04\times {10}^{-24}\mathrm{J}$.

Calculate the number of photons absorbed

Finally, we calculate the number of photons absorbed by dividing the total energy required to heat the water by the energy per photon: $N=\frac{Q}{E}$ where: $N$ = number of photons absorbed $Q$ = heat energy (Joules) $E$ = energy per photon (Joules) Plugging in the values, we get: $N=\frac{15700}{2.040\times {10}^{-24}}$ $N=7.696\times {10}^{27}$ Approximately, $7.70\times {10}^{27}$ photons need to be absorbed to heat the water to boiling point.

Key concepts

Understanding Specific Heat Capacity

Specific heat capacity is an intrinsic property of a material that tells us how much energy must be added, in the form of heat, to raise the temperature of one gram of the substance by one degree Celsius. In the context of our exercise, water has a specific heat capacity of 4.18 J/°C·g. This value is quite high, meaning water can absorb a lot of energy before its temperature rises significantly.

Understanding this concept is crucial when heating substances as it dictates the energy required for a temperature change. For example, in our microwave oven scenario, to calculate the energy needed to bring water to a boil, we can use the formula:

$$Q=mc\mathrm{\Delta }T$$

where Q is the heat energy, m is the mass of water, c is the specific heat capacity, and $\mathrm{\Delta }T$ is the change in temperature. Understanding specific heat capacity allows us to predict the thermal response of water when it absorbs microwave radiation.

Energy Transfer in a Microwave Oven

Energy transfer in a microwave oven involves converting electrical energy into electromagnetic waves that are then absorbed by food or liquids, such as water in our exercise. Microwaves cause water molecules to oscillate, generating heat through this friction and vibration.

To calculate how long it would take to boil water in the microwave, we use the concept of power, which is the rate of energy transfer. We can determine the time required to transfer the necessary amount of energy to the water by using the formula:

$$t=\frac{Q}{P}$$

Here, t is the time, Q is the total energy needed (as calculated using the specific heat capacity), and P is the power output of the microwave. By understanding energy transfer and power, students can solve practical problems involving heat and time, as demonstrated in our given microwave exercise.

Photon Energy Calculation in Microwaves

Microwaves work by emitting photons, which are particles of light that carry energy. To understand how many photons are needed to boil water, we must first calculate the energy of a single photon. Using the equation derived from Planck's discovery and the principles of quantum mechanics:

$$E=\frac{hc}{\lambda }$$

The energy E of a photon is determined by Planck's constant h, the speed of light c, and the photon's wavelength \lambda. In our problem, by knowing the wavelength of the microwave's photons, we can determine the energy per photon.

Then, to find the total number of photons required to boil the water, we divide the total energy needed by the energy per photon. This photon energy calculation is an excellent demonstration of how classical thermodynamics and quantum physics come together in everyday appliances like microwave ovens.