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Heating 235 g of water from 22.6°C to 94.4°C in a microwave oven requires 7.06 × 104 J of energy. If the microwave frequency is 2.88 × 1010 s−1, how many quanta are required to supply the 7.06 × 104 J? The value for Planck's constant is 6.63 × 10-34 Jᐧs/quantum. The formula to use is Energy / (Planck's constant x Frequency).

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Answer: 3.69 × 10^27

Explanation:

Amount of energy required = 7.06 × 10^4 J

Frequency of microwave (f) = 2.88 × 10^10 s−1

Planck's constant (h) = 6.63 × 10^-34 Jᐧs/quantum

Recall ;

Energy of photon = hf

Therefore, energy of photon :

(6.63 × 10^-34)j.s× (2.88 × 10^10)s^-1

= 19.0944 × 10^(-34 + 10) = 19.0944×10^-24 J

Hence, number of quanta required :

(7.06 × 10^4)J / (19.0944 × 10^-24)J

= 0.369 × 10^(4 + 24) = 0.369×10^28

= 3.69 × 10^27

onyebuchinnaji

The number of quanta required to to supply the energy needed to heat up the water is 3.7 x 10²⁷ photons.

The given parameters;

  • mass of the water, m = 235 g
  • initial temperature = 22.6 °C
  • final temperature,  = 94.4 °C
  • energy required to heat up the water, E = 7.06 x 10⁴ J
  • frequency of the microwave, f = 2.88 x 10¹⁰ Hz

The energy of a single photon is calculated as follows;

E = hf

where;

  • h is Planck's constant = 6.63 x 10⁻³⁴ Js
  • f is the frequency of the photon

E = (6.63 x 10⁻³⁴ )(2.88 x 10¹⁰)

E = 1.909 x 10⁻²³ J/photon

The number of quanta required to to supply the energy needed to heat up the water is calculated as follows;

[tex]no \ quanta = \frac{7.06 \times 10^{4}}{1.909\times 10^{-23}} = 3.7 \times 10^{27} \ photons[/tex]

Thus, the number of quanta required to to supply the energy needed to heat up the water is 3.7 x 10²⁷ photons.

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