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Unformatted text preview: 2 We define T m as the temperature between rod 1 and rod 2. Then we have that the heat flows are given by the following equations: Flow out of Coffee pot: Q f = AT 4 f (1) Flow through second part of the rod: Q 2 = k 2 A 2 L 2 ( T m T f ) (2) Flow through first part of the rod: Q 1 = k 1 A 1 L 1 ( T ? T m ) (3) Since we want to keep our coffee at constant temperature: Q f = Q 2 (4) Q 2 = Q 1 (5) From equations (4), (1) and (2), we can solve for T m and get: T m = AT 4 f + k 2 A 2 L 2 T f L 2 k 2 A 2 (6) Similarly from equations (5), (2) and 3, we get: T ? = k 2 A 2 L 2 + k 1 A 1 L 1 T m k 2 A 2 L 2 T f L 1 k 1 A 1 (7) Finally, substituting (6) into equation (7) we get our answer: T ? = k 2 A 2 L 2 + k 1 A 1 L 1 AT 4 f + k 2 A 2 L 2 T f L 2 k 2 A 2 k 2 A 2 L 2 T f L 1 k 1 A 1 (8) 1 4) We assume all the ice melts and all the steam condenses, so we have only water. With m steam = m ice = m , we have heat added to system 1 + heat added to system 2 = 0 (mL s + mc water T sw ) + ( mL i + mc i T iw ) = 0 22.6 10 5 J/kg  (4186 J/kg C)(100C T ) + 3.33 10 5 J/kg + (4186 J/kg C)( T 0C) = 0 which gives T = 280C. Because we can not have all water at this temperature, our initial assumption is wrong. We must have some water and some steam at 100C. If the amount of steam that has condensed is xm , we have heat added to system 1 + heat added to system 2 = 0 xmL s + mL i + mc i T iw = 0 x (22.6 10 5 J/kg) + 3.33...
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 Fall '08
 Packard
 Heat

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