assign4solt12k6 - 4 3: a) The heat current will be the same...

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3: a) The heat current will be the same in both metals; since the length of the copper rod is known, H = (385.0 W/m K)(4.00 x 10 -4 m 2 ) . 39 . 5 ) 00 . 1 ( ) 0 . 35 ( W m K = b) The length of the steel rod may be found by using the above value of H in H = k A D T/L and solving for L 2 , or, since H and A are the same for the rods, L 2 = . 242 . 0 ) 0 . 35 ( ) 0 . 65 ( ) / 0 . 385 ( ) / 2 . 50 ( ) 00 . 1 ( 2 2 m K K K m W K m W m T T k k L = = D D 1: Using H = Ae s T 4 with e = 1, a) (5.67 x 10 -8 W/m 2 K 4 )(273 K) 4 = 315 W/m 2 . b) A factor of ten increase in temperature results in a factor of 10 4 increase in the output; 3.15 x 10 6 W/m 2 . 2: a) The work done by friction is the loss of mechanical energy, ( ) . 10 54 . 1 ) / 50 . 2 ( 2 1 9 . 36 sin ) 00 . 8 )( / 8 . 9 ( ) 0 . 35 ( 2 1 3 2 2 2 2 2 1 J x s m m s m kg v v m mgh o = ˜ ˆ - = - + b) Using the result of part (a) for Q in Q = mc T gives T = (1.54 x 10 3 J)/((35.0 kg)(3650 J/kg K)) = 1.21 x 10 -2 º C.
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4. An ideal monatomic gas with initial volume V 1 = 3 . 50 cm 3 and T 1 = 360 K follows the cycle given. (a) How much work is done by the gas from 1 to 2? (b) How much work is done by the gas from 3 to 4? (c) How much work is done going clockwise around the cycle
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This note was uploaded on 07/20/2009 for the course PHYS physics 10 taught by Professor Goatman during the Spring '08 term at The University of British Columbia.

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assign4solt12k6 - 4 3: a) The heat current will be the same...

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