A well-lagged bar of length 100cm has its ends maintained at 100oC and 40oC respectively. What is the temperature at a point 60cm from the hotter end?
In a well-lagged bar at steady state, temperature decreases linearly with distance
Temperature gradient = \(\frac{\Delta T}{\text{L}}\)
= \(\frac{100 - 40}{100}\) = 0.6º C/cm
Temperature drops at 60cm = 0.6 x 60 = 36º C
Final temperature = 100ºC - 36ºC = 64ºC
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