Microwave ovens have become popular in kitchens everywhere. They are also used in the laboratory, particularly in drying samples for chemical analysis. A typical microwave oven uses radiation with a 12.2 cm wavelength.

a. How many moles of photons of this radiation are required to raise the temperature of 345 g of water from 26.5 ºC to 99.8 ºC?

b. The watt is a unit of power—the rate at which energy is delivered or consumed: 1 W = 1 J/s. Assume that all the energy of a 700. W microwave oven is delivered to the heating of water described in (a). How long will it take to heat the water?

The calculation involves determining the photon energy using Planck's formula, then the energy necessary to heat the water using a specific heat formula. Finally, finding the photons number using Avogadro constant and then calculating time using Power (P) equals to Energy (E) over Time (T).

Explanation:

A microwave oven uses electromagnetic radiation, specifically microwaves, to heat food. To answer the questions, we first need to understand the concepts of electromagnetic radiation, energy, and heat absorption.

Photon energy can be calculated using the formula E=hc/λ, where 'h' is Planck's constant, 'c' is the speed of light, and 'λ' is the wavelength. With a wavelength of 12.2 cm or 0.122 m, the energy of a photon is approximately 1.63x10^-23 Joules.

Next, energy required to heat water is calculated by the formula Q=mcΔT, where 'm' is the mass of water, 'c' is the specific heat of water, and 'ΔT' is the temperature change. For 345 g of water with a temperature change from 26.5°C to 99.8°C, approximately 28634 Joules are required.

Using the Avogadro constant and the energy of a photon, we could determine the moles of photons required and the time required for a 700 W microwave oven to heat the water.

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skyluke89 skyluke89 · Answer 2 · 2024-05-29T10:20:31-04:00

(a) [tex]1.63\cdot 10^5 mol[/tex]

First of all, we need to calculate the energy needed for to raise the temperature of the water:

[tex]Q=m C_S \Delta T[/tex]

where

m = 345 g is the mass of the water

Cs = 4.186 g/J C is the specific heat of water

[tex]\Delta T=99.8C-26.5C=73.3^{\circ}C[/tex] is the change in temperature of the water

Substituting,

[tex]Q=(345 g)(4.186 J/gC)(73.3 C)=1.06\cdot 10^5 J[/tex]

Then we can calculate the energy of one microwave photon with wavelength

[tex]\lambda=12.2 cm=0.122 m[/tex]:

[tex]E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{0.122 m}=1.63\cdot 10^{-24} J[/tex]

where h is the Planck constant and c is the speed of light

So the total number of photons needed is

[tex]n=\frac{Q}{E}=\frac{1.06\cdot 10^5 J}{1.63\cdot 10^{-24} J}=9.8\cdot 10^{28}J[/tex]

And the corresponding number of moles is

[tex]n=\frac{N}{N_A}=\frac{9.8 \cdot 10^{28}J}{6.022\cdot 10^{23} }=1.63\cdot 10^5 mol[/tex]

where [tex]N_A[/tex] is the Avogadro number (the number of photons in 1 mole).

(b) 151.4 s

The power delivered by the microwave oven is

P = 700 W

We know that the power is the ratio of the energy delivered by the time taken:

[tex]P=\frac{E}{t}[/tex]

since in this case the total energy needed to heat the water is [tex]1.06\cdot 10^5 J[/tex], then the time needed is

[tex]t=\frac{E}{P}=\frac{1.06\cdot 10^5 J}{700 W}=151.4 s[/tex]