F06-MT1-Review-Solns

F06-MT1-Review-Solns - 2 We define T m as the temperature...

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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 =- AσεT 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 = AσεT 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 ¶ AσεT 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 Δ T sw ) + ( mL i + mc i Δ T iw ) = 0 -22.6 × 10 5 J/kg - (4186 J/kg · C°)(100°C – T ) + 3.33 × 10 5 J/kg + (4186 J/kg · C°)( T – 0°C) = 0 which gives T = 280°C. Because we can not have all water at this temperature, our initial assumption is wrong. We must have some water and some steam at 100°C. 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 Δ T iw = 0 -x (22.6 × 10 5 J/kg) + 3.33 ×...
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This note was uploaded on 09/09/2008 for the course PHYSICS 7B taught by Professor Packard during the Fall '08 term at Berkeley.

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F06-MT1-Review-Solns - 2 We define T m as the temperature...

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