Question

Suppose that the microwave radiation has a wavelength of 10.8 cm. How many photons are required to heat 205 mL of coffee from 25.0 °C to 62.0°C? Assume that the coffee has the same density, 0.997 g/mL, and specific heat capacity, 4.184 J/( 8K), as water over this temperature range. Express the number of photons numerically. View Available Hint(s) o A¢ ROO? photons

Answer 1

(1)

Mass of coffee = density x volume = 0.997 g/mL x 205 mL = 204.385 g

The amount of heat required = m x c x ▵T = 204.385 g x 4.184 J/g^oC x (62 - 25)^oC = 31640.43 J


Since, Energy = hc/λ = 6.626 x 10^-34 J.s x 3 x 10⁸ m/s / (10.8 x 10^-2 m) = 1.84 x 10^-24 J/photon.

So, the number of photons required = 31640.43 J/1.84 x 10^-24 J/photon = 1.72 x 10²⁸ photons.

(2)

Mass of substance, m = 23 g

Initial temperature, T₁ = 29.0 ^oC

Heat absorbed = 2.30 kJ x 1000 J/ 1 kJ = 2300 J

Final temperature, T2 = ?

Part A: Gold

Specific heat of gold, c = 0.128 J/g°C.

q = m x c x (T₂ - T₁)

where, q = heat , m = mass , c = specific heat , T₁ and T₂ initial and final temperature

Lets put the values in the formula:

2300 J = 23 g x 0.128 J/g°C x (T₂ - 29 ^oC)

781.25 ^oC = T₂ - 29 ^oC

T₂ = 810.25 ^oC

Final temperature of gold = 810.25 ^oC

Part B: Silver

Specific heat of silver, c = 0.235 J/g°C.

q = m x c x (T₂ - T₁)

where, q = heat , m = mass , c = specific heat , T₁ and T₂ initial and final temperature

Lets put the values in the formula:

2300 J = 23 g x 0.235 J/g°C x (T₂ - 29 ^oC)

425.53 ^oC = T₂ - 29 ^oC

T₂ = 454.53 ^oC

Final temperature of silver = 454.53 ^oC

Part C: Aluminium

Specific heat of Aluminium, c = 0.91 J/g°C.

q = m x c x (T₂ - T₁)

where, q = heat , m = mass , c = specific heat , T₁ and T₂ initial and final temperature

Lets put the values in the formula:

2300 J = 23 g x 0.91 J/g°C x (T₂ - 29 ^oC)

109.89 ^oC = T₂ - 29 ^oC

T₂ = 138.89 ^oC

Final temperature of aluminium = 138.89 ^oC

Part D: Water

Specific heat of Water, c = 4.184 J/g°C.

q = m x c x (T₂ - T₁)

where, q = heat , m = mass , c = specific heat , T₁ and T₂ initial and final temperature

Lets put the values in the formula:

2300 J = 23 g x 4.184 J/g°C x (T₂ - 29 ^oC)

23.90 ^oC = T₂ - 29 ^oC

T₂ = 49.90 ^oC

Final temperature of water = 49.90 ^oC