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babymother [125]
3 years ago
13

A microwave oven operates on the principle that application of a high-frequency field causes electrically polarized molecules su

ch as water within the food to oscillate. Rapid alignment and reversal of water molecules results in a nearly uniform generation of thermal energy within the food and, in turn, cooking of the food. When the food is initially frozen, however, the water molecules do not readily oscillate in response to the microwaves, and the volumetric heat generation rates are between one and two orders of magnitude lower than if the water were in liquid form. Microwave power that is not absorbed in the food is reflected back to the microwave generator, where it is dissipated in the form of heat to prevent damage to the generator.
a. Consider a frozen, 1-kg spherical piece of ground beef at an initial temperature of -20 degrees C placed in a microwave oven with total power of 1 kW, air temperature of 30 degrees C, and heat transfer coefficient of 15 W/m^2*K. Assuming the temperature within the beef to be uniform, determine how long it would take the beef to reach a temperature of 0 degrees C, with all the water in the form of ice and 3% of the oven power absorbed by the food. You may assume the physical properties of the beef to be the same as those of ice (p = 920 kg/m^3; c = 2,000 J/kg*K; k = 2.2 W/m*K) in this case.

b. After all the ice is converted to liquid, determine how long it would take to heat the beef to a uniform temperature of 80 degrees C if the temperature within the beef is still assumed to be uniform and 95% of the oven power is absorbed in the food. Assume the properties of the beef are now the same as those of liquid water (p = 1,000 kg/m^3; c = 4,200 J/kg*K; k = 0.6 W/m*K).

c. When thawing food in microwave ovens, you may observe that part of the food is still frozen while other parts of the food are overcooked. Can you explain why this occurs? Explain why most microwave ovens have thaw cycles with very low oven power

Physics
1 answer:
Tema [17]3 years ago
6 0

Answer:

a

The time it would take the beef to reach a temperature of 0⁰C is t= 662 sec = 11.93 min

b)

The time it would take to heat the beef to uniform temperature of 80⁰C is

t = 365 sec =5.93 min

c

 Efficient absorption of Microwave power is in regions of liquid water . Hence if the food or the microwave irradiation  is not uniform, the power will be absorbed non-uniformly, resulting in a non-uniform temperature rise. Defrost region will absorb more energy per unit volume than frozen region.if food is of low thermal conductivity, there will be insufficient time for heat conduction to make the temperature more uniform.Use of low power allows more time for conduction to occur.  

Explanation:

The explanation is shown on the first second and third uploaded image

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The temperature of a substance depends on what two types of energy​
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Thermal and kinetic.

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Thermal energy is the energy generated by heat. Kinetic energy is the movement of particles in a substance which is responsible for the heat increase.

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Which term describes the transfer of thermal energy from one object to another because of a difference in temperature? 1.energy
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The answer is 3.heat
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Use the drop-down menus to complete the statement about Earth’s magnetism.
Nataly [62]

Answer:

The Earth's magnetism is generated in the core, which is composed of iron that is constantly churning

Explanation:

Magnetic fields are produced by charges in motion, therefore by currents.

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5 0
3 years ago
Read 2 more answers
horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothes
lbvjy [14]

Answer:

The tension is  T =  103.96N

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

           The distance between the two poles is D =14 m

          The mass tied between the two cloth line is  m = 3Kg

         The distance it sags is d_s = 1m

The objective of this solution is to obtain the magnitude of the tension on the ends of the  clothesline

Now the sum of the forces on the y-axis is zero assuming  that the whole system is at equilibrium

       And this can be mathematically represented as

                             \sum F_y = 0

 To obtain \theta we apply SOHCAHTOH Rule

 So    Tan \theta = \frac{opp}{adj}

          \theta = tan^{-1} [\frac{opp}{adj} ]

            = tan^{-1} [\frac{1}{7}]

          =8.130^o

=>  \  \ \ \ \ \ \ \ 2T sin\theta -mg =0

=>  \  \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}

=>  \  \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }

=>  \  \ \ \ \ \ \ \  T =\frac{29.4}{2sin(8.130)}

=>  \  \ \ \ \ \ \ \  T = 103.96N

             

                 

5 0
3 years ago
Read 2 more answers
Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and
lawyer [7]

1 kilometre=1000 metre

      1 hour = 3600 second

       1\ km/hr=\frac{1000}{3600} m/s

       1\ km/hr=\frac{5}{18} m/s

The initial velocity of car A is 35.0 km/hr i.e

                                         35.0\ km/hr=35*\frac{5}{18} m/s

                                                                   = 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as  25\ km/hr^2

                                            =\ 25*\frac{1000}{3600*3600} m/s^2

                                            =0.00192901234 m/s^2

The time taken by car A = 15 min.

From equation of kinematics we know that-

                                 v= u+at      [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A,  v = 9.72 m/s +[0.00192901234×15×60]m/s

                                   =11.456111106 m/s

The acceleration of B is given as    15\ km/hr^2

                                    =0.00115740740740 m/s^2

The time taken by car B =20 min

The final velocity of B is -

                             v= u+at

                               = u-at    [Here a is negative due to deceleration]

                               =12.5 m/s +[0.0011574074074×20×60]

                               =13.8888888.....

                               =13.9

The acceleration of C is given as    40\ km/hr^2          

                                                            =\ 0.003086419753 m/s^2

The time taken by car C =30 min

The final velocity of C is-

                                v = u+at

                                   =8.89 m/s+[0.003086419753×30×60] m/s

                                   =14.4455555555..m/s

                                   =14.45 m/s

The car C is decelerating.The deceleration is given as-  60\ km/hr^2

                                                                      =0.0046296296296m/s^2

The time taken by car D= 45 min.

The final velocity of the car D is-

                     v =u+at

                        =30.56 -[0.00462962962962×45×60]m/s

                        =18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

                 


3 0
3 years ago
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