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Thermodynamics

# The temperature gradient in a rod 0.5 m long is 80 c/m . The temperature of the hot end of the rod is $\;30^{0}C\;$. Then the temperature of the cooler and is

$(a)\;40^{0}C\qquad(b)\;10^{0}C\qquad(c)\;-10^{0}C\qquad(d)\;15^{0}C$

Answer : $\;-10^{0}C$
Explanation :
$\large\frac{T_{H}-T_{C}}{L}=\large\frac{\bigtriangleup T}{L}$
$80 = \large\frac{T_{H}-T_{L}}{0.5}$
$T_{L}=T_{H} - 40 = -10^{0}C$