An energy input of 33.44kJ is required to raise the temperature of 100 grams of water from room temperature (20°C) to boiling temperature (100°C). How many photons of energy are needed to bring 250 grams of room temperature water to boiling if 288 mili-meter electro-magnetic radiation is used to heat the water?

1.21 × 10²⁵ photons

Explanation:

If 33.44kJ is required to raise the temperature of 100 grams of water from room temperature (20°C) to boiling temperature (100°C), then the quantity of heat Q 33.44kJ is required to raise the temperature of 250 grams of water from room temperature (20°C) to boiling temperature (100°C) is

Q/33.44 kJ = 250 g/100g

Q = 2.5 × 33.44 kJ = 83.6 kJ

Now the energy of a photon of 288 millimeter electromagnetic radiation is

E = hc/λ where h = planck's constant = 6.63 × 10⁻³⁴ Js, c = speed of light = 3 × 10⁸ m/s and = 288 × 10⁺³ m = 2.88 × 10⁻⁵ m

E = 6.63 × 10⁻³⁴ Js × 3 × 10⁸ m/s ÷ 2.88 × 10⁻⁵ m

= 19.89/2,88 × 10⁻²¹ J

= 6.91 × 10⁻²¹ J

The number of photons of the 288 mm radiation that will raise 250 grams of water from room temperature to boiling point is thus,

n = Q/E = 83.6 kJ/6.91 × 10⁻²¹ J

= 83.6 × 10³ J/6.91 × 10⁻²¹ J

= 12.1 × 10²⁴ photons

= 1.21 × 10²⁵ photons

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