One container holds 0.10 kg of water at 75 ∘C and is warmed to 95 ∘C by heating from contact with the other container. The other container, also holding 0.10 kg of water, cools from 35 ∘C to 15 ∘C. Specific heat of water is 4180 J/kg⋅∘C.Estimate the total change in entropy of two containers of water using the actual temperatures to determine the heat transferred to each container and the average temperatures to determine the change in entropy.Is this energy transfer process allowed by the first law of thermodynamics? Yes or No.is this energy transfer process allowed by the second law of thermodynamics? Yes or No

Respuesta :

Answer:

(1) Total change in Entropy change = [tex]-4.67[/tex] [tex]\frac{J}{K}[/tex]

(2) Yes, energy transfer allowed by the first law of thermodynamics.

(3) No, energy transfer is not allowed by the second law of   thermodynamics.

Explanation:

Given :

For one container

Mass of water [tex]m = 0.10[/tex] Kg

Temperature [tex]T _{1} =[/tex] 75°C

Another Temperature  [tex]T_{2} =[/tex] 95°C

For other container

Mass of water [tex]m = 0.10[/tex] Kg

Temperature [tex]T_{1} =[/tex] 35°C

Another Temperature [tex]T_{2} =[/tex] 15°C

Specific heat of water [tex]C = 4180[/tex] [tex]\frac{J}{Kg C}[/tex]

From the formula of change in entropy in terms temperature,

   [tex]dS = mC \ln \frac{T_{2} }{T_{1} }[/tex]

For one container,

[tex]dS_{1} = 0.1 \times 4180 \times \ln \frac{T_{2} }{T_{1} }[/tex]

Where [tex]T_{2} = 95+273 = 368[/tex] K, [tex]T_{1} = 75 + 273 = 348[/tex] K

[tex]dS_{1} = 23.38[/tex] [tex]\frac{J}{K}[/tex]

For another container,

[tex]dS_{2} = 0.1 \times 4180 \times \ln \frac{T_{2} }{T_{1} }[/tex]

Where [tex]T_{2} = 15 +273 = 288[/tex] K, [tex]T_{1} = 273 + 35 = 308[/tex] K

[tex]dS_{2} = -28.05[/tex] [tex]\frac{J}{K}[/tex]

Total change entropy,

[tex]\Delta S = 23.38 - 28.05 = -4.67[/tex] [tex]\frac{J}{K}[/tex]

Yes, energy transfer allowed by the first law of thermodynamics

No, energy transfer is not allowed by the second law of thermodynamics