Higher 1A T2 2012 Solutions

# 17 105 020 1200 104 53 158 105 hf 12 104 53 hf

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Unformatted text preview: P= # A mg 5 × 9. 8 # Pf = Pi + = 1.17 × 105 + = 1.58 × 105 P a A 12.0 × 10−4 # # \$\$\$"## quickly ⇒ adiabatic P V γ = const # # R + 1fR 5 CP # 2 = = γ= 1 CV 3 # 2fR # 5/3 5/3 Pi V i = Pf V f # 1.17 × 105 × (0.20 × 12.00 × 10−4 )5/3 = 1.58 × 105 × (hf × 12 × 10−4 )5/3 # (hf × 12 × 10−4 )5/3 = 8.8635 × 10−7 # hf × 12 × 10−4 = 2.004 × 10−4 # hf = 16.7cm # # \$&"# P V = const # Pi V i = Pf V f # 1.17 × 105 × 0.20 × 12.00 × 10−4 = 1.58 × 105 × hf × 12 × 10−4 # hf = 14.8cm # dT '"#\$"## P = kA| | dx # P is constant along the rod # (70 − 45) ⇒ P = 385 × 10 × 10−4 × = 9.63W # 1.00 # (45 − 15) \$\$"# 9.625 = 50.2 × 10.0 × 10−4 × L2 # L2 = 0.156m # # ("#\$"## ∆Eint = 0 as isothermal # \$\$"# ∆Eint = Q + W W = 0 as volume does not change # ∆Eint = −233kJ # # \$\$\$"# WC →A = ∆EintC →A − QC →A QC →A = 0 a adiabatic ∆EintC →A = −(∆EintA→B + ∆EintB →C ) as no change...
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## This note was uploaded on 05/31/2013 for the course PHYSIC 1131 taught by Professor John during the Three '13 term at University of New South Wales.

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