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
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
1 answer:
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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Answer:
K.Eₓ = 4 K.E
K.Eₓ = 9 K.E
Explanation:
Th formula for the kinetic energy of a body is given as follows:
---------------equation (1)
where,
K.E = Kinetic Energy of Automobile
m = mass of automobile
v = speed of automobile
For twice speed:
vₓ = 2v
then,
using equation (1):
<u>K.Eₓ = 4 K.E</u>
For thrice speed:
vₓ = 3v
then,
using equation (1):
<u>K.Eₓ = 9 K.E</u>
a) The train collide after 22.5 seconds
b) The trains collide at the location x = 537.5 m
c) See graph in attachment
d) The freight train must have a head start of 500 m
e) The deceleration must be smaller (towards negative value) than
f) The two trains avoid collision if the acceleration of the freight train is at least
Explanation:
a)
We can describe the position of the passenger train at time t with the equation
where
is the initial velocity of the passenger train
is the deceleration of the train
On the other hand, the position of the freight train is given by
where
is the initial position of the freight train
is the constant velocity of the train
The collision occurs if the two trains meet, so
This is a second-order equation that has two solutions:
t = 22.5 s
t = 177.5 s
We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.
b)
In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.
The position of the freight train is
And substituting t = 22.5 s, we find:
We can verify that the passenger train is at the same position at the time of the collision:
So, the two trains collide at x = 537.5 m.
c)
In the graph in attachment, the position-time graph of each train is represented. We have:
- The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
- The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating
The two trains meet at t = 22.5 s, where the position is 537.5 m.
d)
In order to avoid the collision, the freight train must have a initial position of
such that the two trains never meet.
We said that the two trains meet if:
Re-arranging,
Substituting the values for the acceleration and the velocity,
The solution of this equation is given by the formula
The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore
Therefore, the freight train must have a head start of 500 m.
e)
In this case, we want to find the acceleration of the passenger train such that the two trains do not collide.
We solve the problem similarly to part d):
Re-arranging
Substituting,
The solution to this equation is
Again, the two trains never meet if the discriminant is negative, so
So, the deceleration must be smaller (towards negative value) than
f)
In this case, the motion of the freight train is also accelerated, so its position at time t is given by
where is the acceleration of the freight train.
Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,
Re-arranging,
And the solution is
Again, the two trains avoid collision if the discriminant is negative, so
Learn more about accelerated motion:
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Answer:
See attached file
Explanation:
