This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ×...
View
Full
Document
This note was uploaded on 09/09/2008 for the course PHYSICS 7B taught by Professor Packard during the Fall '08 term at Berkeley.
 Fall '08
 Packard
 Heat

Click to edit the document details