1) Find the theoretical delta T
ΔT=K⋅mBA⋅1000MBA⋅mB=4.9⋅2⋅100025⋅122.12=3.21\Delta T=K·\frac{m_{BA}·1000}{M_{BA}·m_{B}}=4.9·\frac{2·1000}{25·122.12}=3.21ΔT=K⋅MBA⋅mBmBA⋅1000=4.9⋅25⋅122.122⋅1000=3.21
2)i=ΔTrealΔT=1.623.21=0.5i=\frac{\Delta T_{real}}{\Delta T}=\frac{1.62}{3.21}=0.5i=ΔTΔTreal=3.211.62=0.5
3)i=1−α(1+n)=1−α(1+0.5)i=1-\alpha (1+n)= 1-\alpha (1+0.5)i=1−α(1+n)=1−α(1+0.5) ⟹ \implies⟹ α=1−i0.5=0.50.5=1\alpha =\frac{1-i}{0.5}=\frac{0.5}{0.5}=1α=0.51−i=0.50.5=1
Answer: 100%
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