Introduction to Health Physics: Fourth Edition - Ruang Baca FMIPA UB

770 CHAPTER 14
(c) The required power to heat the sand–resin mixture from 18 ◦ Cto107 ◦ Cis
P = 0.97 × 500 kg
h × (107◦C − 18 ◦ C) × 1250 J/kg
◦C 1 hour
J
×
= 14,988
3600 seconds s .
(d) Total power dissipated in the cores is
1428 + 9375 + 14,988 = 25,791 J
s ,
and, at 50% efficiency, the required power is
P =
25,791 J/s
0.5
× 1W 1kW
× = 51.6 kW.
J/s 1000 W
Microwave cooking is widely used because of the speed with which food can be
heated in this way. In a conventional oven, the oven’s interior, including the walls and
the air, must be heated. Heat then is transferred from the hot air to the surface of the
food, and it then flows by conduction into the material being cooked. The overall
process is relatively inefficient and slow. In microwave cooking, on the other hand,
the oven and its interior environment are not heated. All the microwave energy is
absorbed by the food. Furthermore, because of the penetration of the microwaves
to a depth of 1–2 cm below the surface of the food, direct deep heating of the
food occurs. The volume of food that must be heated by conduction from the outer
layers is thus greatly diminished. This combination of efficient energy utilization and
deep heating leads to rapid cooking, as shown in Example 14.23. The deep-heating
effect of microwave irradiation is the basis of medical microwave diathermy, in which
beneficial amounts of heat can be applied to inflamed or injured joints and tissues
inside the body without overheating the skin.
W Example 14.23
How long will it take to heat 0.45 kg of fresh peas in 0.05-L water to a temperature of
75 ◦ C from a room temperature of 18 ◦ C in a microwave oven whose microwave power
output is 700 W if the specific heat of the pea-and-water mixture is 3760 J/kg/ ◦ C?
Solution
The required amount of heat energy is
Q = 0.5 kg× 3760 J/kg
◦ C × (75 ◦ C − 18 ◦ C) = 1.1 × 10 5 J.