Answer:
1.21 × 10²⁵ photons
Step-by-step 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