Answer:
W= -2.5 (p₁*0.0012) joules
Explanation:
Given that p₀= initial pressure, p₁=final pressure, Vi= initial volume=0 and Vf=final volume= 6/5 liters where p₁=p₀ then
In adiabatic compression, work done by mixture during compression is
W= 
where f= final volume and i =initial volume, p=pressure
p can be written as p=K/V^γ where K=p₀Vi^γ =p₁Vf^γ
W= 
W= K/1-γ ( 1/Vf^γ-1 - 1/Vi^γ-1)
W=1/1-γ (p₁Vf-p₀Vi)
W= 1/1-1.40 (p₁*6/5 -p₀*0)
W= -2.5 (p₁*6/5*0.001) changing liters to m³
W= -2.5 (p₁*0.0012) joules
<u>Substance</u><u>:</u><u>-</u>
- A substance is matter which has a specific composition and specific properties.
<span>ΔT for the first sample is the total samples final temp, minus the first sample's initial temp (47.9-22.5), so 25.4oC.
Calculating q for the first sample as 108g x 4.18 J/g C x 25.4oC = 11466.58 Joules
Figuring that since the first sample gained heat, the second sample must have provided the heat, so doing the calculation for the second sample, I used
q=mCΔT
11466.58 Joules = 65.1g x 4.18 J / g C x ΔT
11466.58/(65.1gx4.18)=ΔT
ΔT=42.14oC
So, since second sample lost heat, it's initial temperature was 90.04oC (47.9oC final temperature of mixture + 42.14oC ΔT of second sample).</span>
