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Fa06Soln3

# Fa06Soln3 - Physics 7B Fall 2006 Homework 3 Solutions 19.74...

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Physics 7B Fall, 2006 Homework 3 Solutions 19.74 The rate of thermal energy flow is the same for the brick and the insulation: Δ Q / Δ t = A ( T 1 T 2 )/ R eff = A ( T 1 T int )/ R 2 = A ( T int T 2 )/ R 1 , where T int is the temperature at the brick-insulation interface. By equating the first term to each of the others, we have R eff ( T 1 T int ) = R 2 ( T 1 T 2 ), and R eff ( T int T 2 ) = R 1 ( T 1 T 2 ). If we add these two equations, we get R eff ( T 1 T 2 ) = R 1 ( T 1 T 2 ) + R 2 ( T 1 T 2 ), which gives R eff = R 1 + R 2 . This shows the usefulness of the R -value. For the R -value of the brick we have R 1 = L 1 / k 1 = [(4.0 in)/(12 in/ft)](3.28 ft/m)/(0.84 J/s · m · C°)(1 Btu/1055 J)( 3600 s /1 h)(5 C°/9 F°) = 0.69 ft 2 · h · F°/Btu. Thus the total R -value of the wall is R eff = R 1 + R 2 = 0.69 ft 2 · h · F°/Btu + 19 ft 2 · h · F°/Btu = 19.7 ft 2 · h · F°/Btu. The rate of heat loss is Δ Q / Δ t = A ( T 1 T 2 )/ R eff = [(240 ft 2 )(10 F°)/(19.7 ft 2 · h · F°/Btu)](1055 J/Btu)/(3600 s/h) = 36 W . 19.90 ( a ) We find the rate of heat flow through the clothing from Δ Q / Δ t = kA ( Δ T / L ) = (0.025 J/s · m · C°)(1.9 m 2 )[34°C – (– 20°C)]/(0.035 m) = 73 W . ( b ) When the clothing is wet, we have Δ Q / Δ t = kA ( Δ T / L ) = (0.56 J/s · m · C°)(1.9 m 2 )[34°C – (– 20°C)]/(0.0050 m) = 1.1 × 10 4 W . 19.101 . The heat of fusion from the ice forming at the bottom of the ice must be conducted through the ice to the air. When the ice has a thickness x , we have d Q /d t = kA Δ T / x = ρ ( A d x /d t ) L f . We separate the variables and integrate: k ! T " L f d t 0 t = x d x 0 D ; whichgives k ! T t " L f = D 2 2 .

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