A steel bar 10cm long is welded end to end with a copper bar 20 cm long. Each bar has the same cross sectional area of 2cm on a side. The left end of the steel bar is in contact with steam at 100C while the right end of the copper bar is in contact with ice at 0C. What is the temperature at the junction between the 2 bars? What is the total rate of heat conduction?I have 90% done I think, can some one tell me where I am wrong?I used Q/t kA delta T /L So I did it for each of the bars, and I asked my teacher and he said the answer for both should be the same which makes sense but I am not getting matching numbers.For steel: Q (14J/smc)(.2m)(100C-T) / (.1m)For Copper Q (390J/smc)(.2m)(T-0C) / .2mSo am I solving for T? If so then what is Q? I am a little confused having 2 unknowns so if someone can point me in the right direction? Thank you!
For the steel bar Q? k??A??(T_r - T_j)/L? for the copper bar Q? k??A??(T_j - T_l)/L? Heat flow through bars is the same, i.e. Q?Q?Q. To eliminate Q just equate the two expressions: k??A??(T_r - T_j)/L? k??A??(T_j - T_l)/L? since A? A? A (k?/L?)?(T_r - T_j) (k?/L?)?(T_j - T_l) solve for junction temperature T_j T_j [ (k?/L?)?T_r + (k?/L?)?T_l) ] /[ (k?/L?) + (k?/L?) ] [ (14Js??m??°C??/0.1m)?100°C + (390Js??m??°C??/0.2m)?0°C) ] /[ (14Js??m??°C??/0.1m) + (390Js??m??°C??/0.2m) ] 6.7°C To find Q solve the two expressions for their temperature difference: Q?L?/(k??A) T_r - T_j Q?L?/(k??A) T_j - T_l To eliminate T_j add the two equations: Q?L?/(k??A) + Q?L?/(k??A) T_r - T_j + T_j - T_l (Q/A)?( (L?/k?) + (L?/k?) ) T_r - T_l Q A?(T_r - T_l) / ( (L?/k?) + (L?/k?) ) 2cm? ? (100°C - 0°C) / ( (0.1m/14Js??m??°C??) + (0.2m/390Js??m??°C??) ) 0.0002 m? ? 100°C / ( (0.1m/14Js??m??°C??) + (0.2m/390Js??m??°C??) ) 2.612 Js??
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