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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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What is the frequency of a photon of emr with a wavelength of 2.55x10-3m?
olga nikolaevna [1]

Answer:

f = 1.18 x 10¹¹ Hz

Explanation:

The equation used to find frequency is:

f = c / w

In this form, "f" represents the frequency (Hz), "c" represents the speed of light (3.0 x 10⁸ m/s), and "w" represents the wavelength (m).

Since you have been given the value of the constant (c) and wavelength, you can substitute these values into the equation to find frequency.

f = c / w                                                      <---- Formula

f = (3.0 x 10⁸ m/s) / w                                 <---- Plug 3.0 x 10⁸ in "c"

f = (3.0 x 10⁸ m/s) / (2.55 x 10⁻³ m)            <---- Plug 2.55 x 10⁻³ in "w"

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3 0
1 year ago
Desde una altura de 120 m se deja caer un cuerpo. Calcular a los 2,5 s i) la rapidez que lleva; ii) cuánto ha descendido; iii) c
stira [4]

Answer:

i) 24.5 m/s

ii) 30,656 m

iii) 89,344 m

Explanation:

Desde una altura de 120 m se deja caer un cuerpo. Calcule a 2.5 s i) la velocidad que toma; ii) cuánto ha disminuido; iii) cuánto queda por hacer

i) Los parámetros dados son;

Altura inicial, s = 120 m

El tiempo en caída libre = 2.5 s

De la ecuación de caída libre, tenemos;

v = u + gt

Dónde:

u = Velocidad inicial = 0 m / s

g = Aceleración debida a la gravedad = 9.81 m / s²

t = Tiempo de caída libre = 2.5 s

Por lo tanto;

v = 0 + 9.8 × 2.5 = 24.5 m / s

ii) El nivel que el cuerpo ha alcanzado en 2.5 segundos está dado por la relación

s = u · t + 1/2 · g · t²

= 0 × 2.5 + 1/2 × 9.81 × 2.5² = 30.656 m

iii) La altura restante = 120 - 30.656 = 89.344 m.

6 0
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