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Norma-Jean [14]
3 years ago
14

Two Carnot air conditioners, A and B, are removing heat from different rooms. The outside temperature is the same for both rooms

, 317 K. The room serviced by unit A is kept at a temperature of 293 K, while the room serviced by unit B is kept at 296 K. The heat removed from either room is 4330 J. For both units, find (a), (b) the magnitude of the work required and (c), (d) the magnitude of the heat deposited outside for A and B conditioners respectively.
Physics
1 answer:
amid [387]3 years ago
8 0

Answer:

a) Work required for air conditioner A = 354.7 J

b) Work required for air conditioner B = 310.3 J

c) The magnitude of the heat deposited outside for conditioner A = 4684.7 J

d) The magnitude of the heat deposited outside for conditioner B = 4640.3 J

Explanation:

In a carnot air conditioner, it operates like a reverse carnot engine; i.e. it removes heat from the cold reservoir (making it colder) and dumps the heat in the hot reservoir (making it hotter)

For a Carnot air conditioner,

Q꜀ is the heat removed from the colder reservoir = 4330 J for both cases

T꜀ is the temperature of the colder reservoir (temperature of the rooms) = 293 K and 296 K for both cases to be considered.

Qₕ is the heat deposited in the warmer reservoir = ? for both cases

Tₕ is the temperature of the hot reservoir (temperature of outside) = 317 K for both cases.

For Carnot air conditioners,

Qₕ = W + Q꜀ (eqn 1)

And

(Qₕ/Tₕ) - (Q꜀/T꜀) = 0 (eqn 2)

Making Qₕ the subject of formula in eqn 2

Qₕ = Tₕ (Q꜀/T꜀)

Substituting this into eqn 1

Tₕ (Q꜀/T꜀) = W + Q꜀

Q꜀ (Tₕ/T꜀) - Q꜀ = W

Q꜀ [(Tₕ - T꜀)/T꜀ ] = W

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ]

For the air conditioner A,

T꜀ = 293 K, Tₕ = 317 K, Q꜀ = 4330 J, W = ?

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ] = 4330 [ (317 - 293)/293] = 354.7 J

For the air conditioner B,

T꜀ = 296 K, Tₕ = 317 K, Q꜀ = 4330 J, W = ?

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ] = 4330 [ (317 - 296)/296] = 310.3 J

c) Qₕ = W + Q꜀

For conditioner A,

Qₕ = 354.7 + 4330 = 4684.7 J

For conditioner B,

Qₕ = 310.3 + 4330 = 4640.3 J

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Answer:

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Explanation:

Specific heat is defined as the amount of energy per unit mass needed to raise the temperature of the substance by a one degree Celsius.

Heat energy gain or loss by any substance is given by :

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Here m is the mass of the substance, C is specific heat of the substance and T₁ and T₂ are the initial and final temperature of the substance.

According to the problem,

Mass of water, m = 190 gm

Specific heat of water, C = 4.186 J gm⁻¹ ⁰C⁻¹

Initial temperature, T₁ = 42.7⁰ C

Final temperature, T₂ = 21.2⁰ C  

Substitute these values in equation (1).

Heat energy loss by water = 190 x 4.186 x ( 21.2 - 42.7 )

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Answer:

-\frac{1}{2}

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Explanation:

Given parameters:

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Unknown:

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Estimate how long a 2500 W electric kettle would take to boil away 1.5 Kg of water . The specific latent heat of vaporization of
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The time it would take a 2500 W electric kettle to boil away 1.5 Kg of water is 2400 seconds

<h3>How to calculate the time</h3>

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Substitute the values into the formula

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Make 'time' the subject

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Therefore, the time it would take a 2500 W electric kettle to boil away 1.5 Kg of water is 2400 seconds.

Learn more about specific latent heat of vaporization:

https://brainly.in/question/1580957

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5 0
3 years ago
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PLEASE HELP ASAP
aleksandrvk [35]

Answer:

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