One end of a copper poker is placed in a fire with temperature of 600°C, while the other end is kept at room temperature (25°). The poker is 1,5 m long and has a radius of 3 x 10-3 m.
Answer
In this case,
"\\frac{dQ}{dt}=kA(T_2-T_1 )\/d\\\\\n\nA=\u03c0r^2"
The amount of heat conducted from one end of the poker to the other in 5.0s is given
"Q=\\frac{dQ}{dt }\u2206t\\\\=((79.5)*\u03c0*(0.005)^2 (502-26))\/1.2 (5)\\\\=12 J."
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