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IRISSAK [1]
3 years ago
7

Se tienen 500g de alcohol etílico a una temperatura de -40 °C ¿Cuánto calor se necesita para transformarlo a vapor a una tempera

tura de 150ºC?
Physics
1 answer:
Feliz [49]3 years ago
8 0

The question is: You have 500g of ethyl alcohol at a temperature of -40 ° C. How much heat is needed to transform it into steam at a temperature of 150ºC?

Answer: 233700 J heat is needed to transform ethyl alcohol into steam at a temperature of -40^{o}C to 150^{o}C.

Explanation:

Given: Mass = 500 g

Initial temperature = -40^{o}C

Final temperature = 150^{o}C

The standard value of specific heat of ethyl alcohol is 2.46 J/g ^{o}C.

Formula used to calculate the heat energy is as follows.

q = m \times C \times (T_{2} - T_{1})

where,

q = heat energy

m = mass of substance

C = specific heat

T_{1} = initial temperature

T_{2} = final temperature

Substitute the values into above formula as follows.

q = m \times C \times (T_{2} - T_{1})\\= 500 g \times 2.46 J/g^{o}C \times [150 - (-40)]^{o}C\\= 233700 J

Thus, we can conclude that 233700 J heat is needed to transform ethyl alcohol into steam at a temperature of -40^{o}C to 150^{o}C.

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Answer:

\beta = 41.68°

Explanation:

according to snell's law

\frac{n_w}{n_g} = \frac{sin\alpha}{sin30 }

refractive index of water n_w is 1.33

refractive index of glass  n_g  is 1.5

sin\alpha = \frac{n_w}{n_g}* sin30

sin\alpha = 0.443

now applying snell's law between air and glass, so we have

\frac{n_g}{n_a} = \frac{sin\alpha}{sin\beta}

sin\beta = \frac{n_g}{n_a} sin\alpha

\beta = sin^{-1} [\frac{n_g}{n_a}*sin\alpha]

we know that sin\alpha = 0.443

\beta = 41.68°

7 0
3 years ago
A 53.0-kg skater is traveling due east at a speed of 2.90 m/s. A 72.0-kg skater is moving due south at a speed of 6.20 m/s. They
soldier1979 [14.2K]

Answer:

a.\thta=71^{\circ}

b.v_f=3.78 m/s

Explanation:

We are given that

m_1=53 kg

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a.We have to find the angle

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\theta=71^{\circ}

b. We have to find the speed v_f

According to law of conservation of momentum

m_1v_1=(m_1+m_2)v_fcos\theta

53(2.9)=(53+72)v_fcos 71^{\circ}=40.7 v_f

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3 years ago
Which formula can be used to solve problems related to the first law of thermodynamics?
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Explanation:

First law of thermodynamics states that the total energy of the system remains conserved. Energy can neither be destroyed, nor be created but it can only be transformed into one form to another.

Its implication is any change in the internal energy will be either due to heat energy or work energy.

Mathematically,

\Delta U=Q+W

where, Q = heat energy

W = work energy

\Delta U = Change in internal energy

Sign convention for these energies:

For Q: Heat absorbed will be positive and heat released will be negative.

For W: Work done by the system is negative and work done on the system is positive.

For \Delta U: When negative, internal energy is decreasing and when positive, internal energy is increasing.

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8 0
3 years ago
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Vaselesa [24]

Answer:B)439.21 Hz

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Given

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