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sammy [17]
3 years ago
12

An 80.0 g sample of a gas was heated from 25 ∘C25 ∘C to 225 ∘C.225 ∘C. During this process, 346 J of work was done by the system

and its internal energy increased by 6085 J.6085 J. What is the specific heat of the gas?
Physics
1 answer:
Ne4ueva [31]3 years ago
6 0

Answer:

402 J/(kg ⁰C)

Explanation:

m = mass of the sample = 80.0 g = 0.080 kg

T₀ = Initial temperature of the sample = 25 ⁰C

T = Final temperature of the sample = 225 ⁰C

W = Amount of work done by the system = 346 J

U = Increase in internal energy = 6085 J

Q = Amount of heat given to the gas

Amount of heat is given as

Q = U + W

Q = 6085 + 346

Q = 6431 J

Heat received by sample is given as

Q = m c (T - T₀ )

inserting the values

6431 = (0.080) c (225 - 25)

c = 402 J/(kg ⁰C)

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The equilibrium temperature is T13=3.12 ◦C

<u>Explanation:</u>

<u>Given </u>

The temperature of liquids: T1=6◦C, T2=23◦C, T3=38◦C

The temperature of 1+2 liquids mix: T12= 13◦C.

The temperature of 2+3 liquids mix: T23=26.8 ◦C.

The temperature of 1+3 liquids mix: T13= ??

<u>1.When the first two liquids are mixed:</u>

  • mC1(T1-T12)+mC2(T2-T12)=0
  • C1(6-13)=C2(23-13)=0
  • 7C1=10C2
  • C1=1.42C2

<u>2.When the second and third liquids are mixed</u><u>:</u>

  • mC2(T2-T23)+mC3(T3-T23)=0
  • C2(23-26.8)=C3(38-26.8)=0
  • 3.8C2=12.8C3
  • C2=3.36C3

<u>3.When the first and third liquids are mixed:</u>

  • mC1(T1-T13)+mC3(T3-T13)=0
  • C1(6-T13)+C3(38-T13)=0
  • C1=1.42C2  C2=3.36C3
  • C1=1.42C2(3.36C3)
  • C1=4.77C3
  • C1(6-T13)+C3(38-T13)=0
  • 4.77C3(6-T13)+C3(38-T13)=0
  • By solving the equation we get,
  • T13=3.12 ◦C
  • The equilibrium temperature is T13=3.12 ◦C

<u></u>

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