ch18-p056 - 56. (a) The surface area of the cylinder is...

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56. (a) The surface area of the cylinder is given by 22 2 2 2 2 2 11 1 1 2 2 2 (2.5 10 m) 2 (2.5 10 m)(5.0 10 m) 1.18 10 m Arr h ππ π −− =+ = × + × × = × , its temperature is T 1 = 273 + 30 = 303 K, and the temperature of the environment is T env = 273 + 50 = 323 K. From Eq. 18-39 we have () ( ) 44 2 2 4 4 e n v 0.85 1.18 10 m (323K) (303K) 1.4W. PA T T σε =− = × −= (b) Let the new height of the cylinder be h 2 . Since the volume V of the cylinder is fixed, we must have Vr hr
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