± Using Microwave Radiation to Heat Coffee Microwave ovens use microwave radiati

Question

± Using Microwave Radiation to Heat Coffee

Microwave ovens use microwave radiation to heat food. The microwaves are absorbed by the water molecules in the food, which is transferred to other components of the food. As the water becomes hotter, so does the food.

Part A

Suppose that the microwave radiation has a wavelength of 11.2 cm . How many photons are required to heat 305 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/(gK) , as water over this temperature range.

Express the number of photons numerically.

Hints

2.94•1028

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± Using Microwave Radiation to Heat Coffee

Microwave ovens use microwave radiation to heat food. The microwaves are absorbed by the water molecules in the food, which is transferred to other components of the food. As the water becomes hotter, so does the food.

Part A

Suppose that the microwave radiation has a wavelength of 11.2 cm . How many photons are required to heat 305 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/(gK) , as water over this temperature range.

Express the number of photons numerically.

Hints

2.94•1028

SubmitMy AnswersGive Up

Explanation / Answer

Heat required to raise the temperature of coffee from 25 to 65 is q

q = m × T × C

Where,

m = mass of Coffee, 304.09g

T = temperature difference ,62 - 25 = 37

C = heat capacity of Coffee , 4.184J/g

q = 304.09g × 37 × 4.184J/g

= 47075J

Energy of radiation ,E = hv

h= Planck constant , 6.626×10^-34J.s

v = frequency = speed of light /Wavelength

v = 2.99× 10^8(m/s)/0.112m

= 2.67×10^9s^-1

Energy of photon , E = 6.626×10^-34J.s × 2.67×10^9s^-1

= 1.77× 10^-24J

No of photons required = 47075 J /1.77×10^-24J

= 2.66×10^28photons

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