A long rod is in perfect thermal contact with boiling water (atatmospheric press

Question

A long rod is in perfect thermal contact with boiling water (atatmospheric pressure) at one end and with a mixture of ice water atthe other end. The rod consists of a 1.00 m copper section (one endof it in boiling water) joined at the other end to lengthL2 of steel (with one end of it in the mixture of icewater). The cross-sectional areas of both sections of the rod is4.00 cm2. The temperature at the copper-steel junctionis 65.0C after a steady state has been set up. (a) How much heatper second flows from the boiling water to the ice water? (b) Whatis the length L2 of the steel section?

Explanation / Answer

Tboiling = 100 0C
Tice = 0 0C
Tmiddle = 65.0 C
Lcopper = 1.00 m
A = 4.00 cm2
   = 4*10-4 m2
kcopper = 385.0 W/m*K
ksteel = 50.2 W/m*K
(a)   the heat current, H, of each section of the rod isequal once a steady state has been set up. Consequently, to find the heat flow, we can simply find theheat flow through the copper portion is
                        H = kcopperA(Tboil - Tmiddle) /Lcopper                             =385.0 * 4*10-4 * (100 - 65.0) / 1.00                            = 5.39 W
heat flow through the rod in one second is 5.39 W * 1 s =5.39 J

(b) The length of the steel rod is             H = ksteelA(Tmiddle - Tice)/ Lsteel
         5.39 = 50.2 *4*10-4 * (65.0 - 0) / Lsteel
       Lsteel = 50.2* 4*10-4 * (65.0 - 0) / 5.39                 =---------- Solve it I hope it helps you

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