Question
Choose the correct relation, when the temperature of an isolated black body falls from $T_{1}$ to $T_{2}$ in time ' $t$ ', and assume ' $c^{\prime}$ to be a constant :(a) $t-c\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)$(b) $t=c\left(\frac{1}{T_{2}^{2}}-\frac{1}{T_{1}^{2}}\right)$(c) $t=c\left(\frac{1}{T_{2}^{3}} \ldots \frac{1}{T_{1}{ }^{3}}\right)$(d) $t=c\left(\frac{1}{T_{2}{ }^{4}}-\frac{1}{T_{1}{ }^{4}}\right)$
Step 1
Mathematically, it can be written as: \[E = \sigma T^{4}\] where \(E\) is the rate of emission of radiation, \(\sigma\) is the Stefan-Boltzmann constant, and \(T\) is the absolute temperature of the black body. Show more…
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The temperature of an isolated black body falls from T1 to T2 in time t. The value of t is (Let c be a constant): A. t = c (1/T2 - 1/T1) B. t = c (1/T2^2 - 1/T1^2) C. t = c (1/T2^3 - 1/T1^3) D. t = c (1/T2^4 - 1/T1^4)
The temperature T = T(x, y) in °C at point (x, y) satisfies Tx(1, 2) = 3 and Ty(1, 2) = -1. If x = e^(2t-2) cm and y = 2 + ln t cm, find the rate at which the temperature T changes when t = 1 sec.
The temperature $T=T(x, y) \quad$ in $\quad^{\circ} \mathrm{C} \quad$ at $\quad$ point $\quad(x, y)$ satisfies $T_{1}(1,2)=3$ and $T_{1}(1,2)=-1 .$ If $x=e^{2 t-2} \mathrm{cm}$ and $y=2+\ln t \mathrm{cm},$ find the rate at which the temperature $T$ changes when $t=1$ sec.
Partial Derivatives
The Chain Rule
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