The answer & explanation for this question is given in the attachment below.
Answer:
ΔS=0.148 KJ/K
Explanation:
Given that
Q = 100 KJ
T₁=200°C
T₁=200+273 = 437 K
T₂=5°C
T₂=5 + 273 = 278 K
Reservoir 1 is rejecting heat that is why it taken as negative while the reservoir 2 is gaining the heat that is why it is taken as positive.
So the total change in entropy given as
ΔS= - Q/T₁ + Q/T₂
ΔS= - 100/473 + 100/278 KJ/K
ΔS=0.148 KJ/K
Answer:
It depends what other values you have. Can you give more info? If they give density then you can solve for m.
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Answer:
The answers to the questions are as follows
- The temperature of the water bath goes up
- The piston moves out
- Heat flows out of the gaseous mixture
- 173.kJ flows out of the system
Explanation:
- From the question, it is noted that 173.kJ of heat flows out of the system into the insulated water bath therefore the temperature of the water bath goes up
as seen in the relation ΔH = m·c·ΔT
Where ΔH heat measured by temperature rise ΔT of a given mass of water m of specific heat capacity of 4.2 J/g°C
- The amount of heat measured from previous experiment is more than the heat from the present reaction therefore since in the present reaction is constant pressure and from the first law of thermodynamics Energy can neither be created nor destroyed, the balance heat will be transformed to work evidence in the piston moving out
ΣH = Q + W where
w = P × ΔV = P × (P₂ - P₂)
- Heat flows out of the gaseous mixture and is sensed from the rise in the temperature of the water bath
- 173.kJ flows out of the system
