Please help I’m torn between A and B. I’ve asked many and people say A, but I asked my science teacher and she said B. And also
I’ve researched and on answers.com also says Water is the better insulator..but idk I’m conflicted! APEX
1 answer:
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The given mass of cobalt chloride hydrate = 2.055 g
A sample of cobalt chloride hydrate was heated to drive off waters of hydration and the anhydrate was weighed.
The mass of anhydrous cobalt chloride = 1.121 g anhydrate.
The mass of water lost during heating = 2.055 g - 1.121 g = 0.934 g
Converting mass of water of hydration present in the hydrate to moles using molar mass:
Mass of water = 0.934 g
Molar mass of water = 18.0 g/mol
Moles of water =
This question is incomplete, the complete question is;
Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar, 520 K. If the carbon dioxide behaves as an ideal gas, determine the amount of entropy produced, in kJ/K. Assuming;
a) constant specific heats Cp = 0.939 kJ/Kg K
b) variable specific heats
Answer:
a) the amount of entropy produced is 0.731599 kJ/K
b) the amount of entropy produced is 0.69845 kJ/K
Explanation:
Given the data in the question;
5 kg of carbon dioxide (CO₂) gas undergoes a process in a well-insulated piston-cylinder assembly.
m = 5 kg
Molar mass M = 44.01 g/mol
P₁ = 2 bar, P₂ = 20
T₁ = 280 K, P₂ = 520 K
Since its insulated { q = 0 } ( kinetic and potential energy effects = 0 )
Now,
a) the amount of entropy produced, in kJ/K, Assuming constant specific heats with Cp = 0.939 kJ/Kg K
S = m × ((Cp × In( T₂/T₁) - R × In( P₂/p₁ ))
we substitute
S = 5 × (( 0.939 × In( 520/280) - 0.1889 × In( 20/2 ))
= 5 × ( 0.5812778 - 0.434958 )
= 5 × 0.1463198
= 0.731599 kJ/K
Therefore, the amount of entropy produced is 0.731599 kJ/K
b) the amount of entropy produced, in kJ/K, Assuming variable specific heats.
Now, from Table A-23: Ideal Gas Properties of Selected Gases;
T₁,T₂ : s₁⁰ = 211.376 kJ/kmol-K, s₂⁰ = 236.575 kJ/kmol-K
now, s₁ = s₁⁰ / M and s₂ = s₂⁰ / M
we substitute
s₁ = s₁⁰ / M = 211.376 / 44.01 = 4.8029 kJ/kg
s₂ = s₂⁰ / M = 236.575 / 44.01 = 5.37548 kJ/kg
S = m × (( s₂ - s₁ ) - R × In( p₂ / p₁ ))
we substitute
S = 5 × (( 5.37548 - 4.8029 ) - 0.1880 × In( 20 / 2 ))
= 5 × ( 0.57258 - 0.432885997 )
= 5 × 0.13969
= 0.69845 kJ/K
Therefore, the amount of entropy produced is 0.69845 kJ/K