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Oksana_A [137]
3 years ago
11

A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a lar

ge tub of ice and water. In 5 minutes of operation, the heat rejected by the engine melts 0.0400 kg of ice. During this time, how much work W is performed by the engine?
Physics
2 answers:
horrorfan [7]3 years ago
7 0

Answer:

W= 4.89 KJ

Explanation:

Lets take

temperature of hot water T₁ = 100⁰C

T₁ = 373 K

Temperature of cold ice T₂= 0⁰C

T₂ = 273 K

The latent heat of ice LH= 334 KJ

The heat rejected by the engine Q= m .LH

Q₂= 0.04 x 334

Q₂= 13.36 KJ

Heat gain by engine = Q₁

For Carnot engine

Q_1=\dfrac{T_1}{T_2}Q_2

Q_1=\dfrac{373}{273}\times 13.36

Q₁  = 18.25 KJ

The work W= Q₁  - Q₂

W= 18.25 - 13.36 KJ

W= 4.89 KJ

vodomira [7]3 years ago
3 0

Answer:

W = 4893.77 J

Explanation:

given,

mass of the ice = 0.04 Kg

time of operation = 5 minutes

Heat delivered  =

  Q_c = m_c \Delta H_f

  Δ H_f is latent heat of fusion of ice = 334 x 10³ J/Kg

   Q_c = 0.04 \times 334 \times 10^3

  Q_c = 13360\ J

Work done

W = Q_H - Q_C

Q_H= (\dfrac{T_H}{T_C})Q_C

W= (\dfrac{T_H}{T_C}-1)Q_C

W= (\dfrac{373}{273}-1)\times 13360

W = 4893.77 J

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The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

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