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lilavasa [31]
3 years ago
10

Even at rest, the human body generates heat. The heat arises because of the body's metabolism - that is, the chemical reactions

that are always occurring in the body to generate energy. In rooms designed for use by large groups, adequate ventilation or air conditioning must be provided to remove this heat. Consider a classroom containing 138 students. Assume that the metabolic rate of generating heat is 128 W for each student and that the heat accumulates during a 60-minute lecture. In addition, assume that the air has a molar specific heat of CV = 5/2R and that the room (volume = 1190 m^3, initial pressure = 1.01 x 10^5 Pa, and initial temperature = 21.0 °C) is sealed shut. If all the heat generated by the students were absorbed by the air, by how much would the air temperature rise?
Physics
1 answer:
ser-zykov [4K]3 years ago
8 0

Answer:

The  value  is   \Delta  T  = 62.2 \ K

Explanation:

From the question we are told that

   The number of students is  N  =  138

   The metabolic rate for each student is   r =  128 W

   The time duration is  t =  60 \  minutes  =  3600 \  s

   The  molar specific heat of air is  C_V =\frac{ 5}{2} R

   The volume  is V  =  1190 \  m^3

    The  pressure is   P  =  1.01 *10^{5} \  Pa

     The  initial temperature is T_i  =  21^o C =  294

Generally the metabolic rate of the students is

              K =  N * r

=>           K =  138  *128

=>           K = 17664 \  W

The  total heat generated by the students is  

             H  =  K  *  t

=>           H  =  17664  * 3600

=>           H  =6.3590*10^{7} \ J

From the ideal gas law  we can evaluate n (number of moles ) as  

           n  =  \frac{PV}{RT_i}

Here  R is the gas constant with value  R =  8.314  \ J / mol . K

So

        n  =  \frac{1.01 *10^{5}  *  1190 }{8.314  *  294}

=>   n  = 49171.2 \  moles

Generally the heat generated is mathematically represented as      

       H  =  n *  C_V  *  \Delta T

=>     H  =  n *  (\frac{ 5}{2} *R)*  \Delta T

=>    \Delta  T  =  \frac{2  * H}{n * 5 *R  }

=>     \Delta  T  =  \frac{2   * 6.3590*10^{7}}{49171.2 * 5 * 8.314}

=>    \Delta  T  = 62.2 \ K

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Please help with these questions as well! I need urgent help! I will give brainliest! God bless!
Radda [10]

6.  Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.  

F = ma

F = (15.3)(-9.8)

F = -150N

Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object.  So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.


7.  Same idea as question 2.  

First determine the weight of the object:

F = ma

F = (30)(-9.8)

F = -294N

The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.

-294 + 150N + x = 0

x = 144N  

So the person is exerting 144 N.


10.  First find the force of block B to the right due to its acceleration:

F = ma

F = (24)(0.5)

F = 12N

So block B is moving 12N to the right relative to block A due to block A's movement to the left.  However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A.  The force that is causing block B to experience the lower relative force to the right is because of the friction.  To find the friction:

The sum of the forces in the leftward and rightward direction for block B must equal 12N.

75 - x = 12

x = 63N

So the force of friction of block A on block B is 63N to the left.


5 0
3 years ago
What is the efficiency of the<br> machine if the input is 263 J and<br> the output is 142 J?
Papessa [141]

Answer: 57.79%

Explanation: 152J/263J=.577946768 or 57.79% or roundedthe nearest whole percent is 58%

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Answer:

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sattari [20]

Answer:

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andriy [413]

Answer:

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Explanation:

Hope This Helps!!

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