Answer:
ΔT = 13.65° C
ΔQ = 13.7 J
Explanation:
First we will find the final temperature of air by using equation of state:
P₁V₁/T₁ = P₂V₂/T₂
For Isochoric Process, V₁ = V₂
Therefore,
P₁/T₁ = P₂/T₂
T₂ = P₂T₁/P₁
where,
T₂ = Final Temperature = ?
P₂ = Final Pressure = 1050 mb
P₁ = Initial Temperature = 1000 mb
T₁ = Initial Temperature = 0°C = 273 k
Therefore,
T₂ = (1050 mb)(273 K)/(1000 mb)
T₂ = 286.65 K
Change in Temperature = ΔT = T₂ - T₁
ΔT = 286.65 K - 273 K
<u>ΔT = 13.65° C</u>
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The first law of thermodynamics can be written as:
ΔQ = ΔU + W
where,
ΔQ = heat absorbed
ΔU = change in internal energy = mCΔT
W = Work Done = 0 (in case of isochoric process)
Therefore.
ΔQ = mCΔT
where,
m = mass of air = 1 g = 1 x 10⁻³ kg
C = specific heat of dry air = 1003.5 J/kg.°C
Therefore,
ΔQ = (1 x 10⁻³ kg)(1003.5 J/kg.°C)(13.65°C)
<u>ΔQ = 13.7 J</u>
Answer:
79%
Explanation:
Energy expelled / Energy absorbed = 415/525 = 0.79 = 79%
