Dt we equate the two expressions for pcond and solve

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Unformatted text preview: the two expressions for Pcond and solve for dh/dt: dh k ( TH − TC ) = . dt LF ρ h Since 1 cal = 4.186 J and 1 cm = 1 × 10–2 m, the thermal conductivity of ice has the SI value k = (0.0040 cal/s·cm·K) (4.186 J/cal)/(1 × 10–2 m/cm) = 1.674 W/m·K. The density of ice is ρ = 0.92 g/cm3 = 0.92 × 103 kg/m3. Thus, ( 1.674 W m ×K ) ( 0°C + 10°C ) dh = = 1.1×10−6 m s = 0.40 cm h . 3 3 3 dt 333 × 10 J kg 0.92 × 10 kg m ( 0.050 m ) ( )( ) _______________________________________________________________________ _...
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This note was uploaded on 05/14/2013 for the course MATH 346 taught by Professor Professormiguelarcones during the Spring '08 term at Binghamton.

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