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klemol [59]
3 years ago
12

A 19.0 g sample of liquid methane is heated at a constant pressure of 1 atm from a temperature of 109.1 K to a temperature of 18

5.3 K. How much energy in kJ is required?
Physics
1 answer:
iragen [17]3 years ago
6 0

Answer:

The energy required is 12.887KJ

Explanation:

There are two separate heat inputs involved in this problem:

  • q₁ = heat added to vaporize the methane at 109.1 K
  • q₂ = energy added to heat the vapor from 109.1 K to 185.3 K

q = q₁ +q₂

q = nΔH + mCΔT

where;

n is number of moles of methane

ΔH is the molar enthalpy of vaporization of methane =8.17 kJ/mol

m is the mass of methane = 19g

C is the specific heat capacity of gaseous methane = 2.20 J/g.K

ΔT = T₂ - T₁ = 185.3 - 109.1 = 76.2 K

n = Reacting mass/Molar mass

molar mass of methane (CH₄) = 16g/mol

n = 19/16 = 1.1875 mol

⇒q₁ = nΔH = 1.1875 X 8.17 = 9.702 kJ

⇒q₂ = mCΔT, = 19 X 2.2 X 76.2 = 3185.16 J = 3.18516KJ

q = q₁ +q₂, ⇒ 9.702 kJ + 3.18516KJ = 12.887KJ

Therefore, the energy required is 12.887KJ

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CAR 2

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M = 2100/30

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8 0
3 years ago
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Answer:

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Explanation:

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The water flowing through a 2.0 cm (inside diameter) pipe flows out through three 1.3 cm pipes. (a) If the flow rates in the thr
Verdich [7]

Answer:

a)54L/min

b)0.845

Explanation:

a) A x V=A_1V_1+ A_2V_2+A_3V_3

where suffix 1,2,3 refers to the three pipes.

            =27L/min+16L/min+11 L/min

            =54L/min

b) A x V=54L/min => \frac{\pi }{4} d^2 x v

   d= 2 cm

\frac{\pi }{4} d^2 x v = 54

v= \frac{4}{\pi } x \frac{54}{2^2}

-> A_1 x V_1=27L/min => \frac{\pi }{4} d_1^2 x v_1

d_1= 1.3cm

\frac{\pi }{4} d^2 x v_1 = 27

v_1= \frac{4}{\pi } x \frac{27}{1.3^2}

Next is to find the ratio of speed i.e \frac{v}{v_1}

\frac{4}{\pi } x \frac{54}{2^2} / \frac{4}{\pi } x \frac{27}{1.3^2} => \frac{54}{27} \frac{1.3^2}{2^2}

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8 0
3 years ago
A steam engine absorbs 4 x 105 J and expels 3.5 x 105 J in each cycle. What is its efficiency?
Wewaii [24]
Input heat, Qin = 4 x 10⁵ J
Output heat, Qout = 3.5 x 10⁵ J

From the first Law of thermodynamics, obtain useful work performed as
W = Qin  -  Qout
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By definition, the efficiency is
η = W/Qin
   = 100*(0.5 x 10⁵/4 x 10⁵)
   = 12.5%

Answer: The efficiency is 12.5%
3 0
3 years ago
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