The temperature dependence of resistance is given by:
\[R = R_0 (1 + \alpha \Delta T).\]
From $0^\circ \text{C}$ to $100^\circ \text{C}$:
\[\frac{\Delta R}{R_0} = \alpha \Delta T \implies \alpha = \frac{10.2 - 10}{10 \cdot 100} = 0.002.\]
From $0^\circ \text{C}$ to $t^\circ \text{C}$:
\[\frac{\Delta R}{R_0} = \alpha \Delta T \implies \Delta T = \frac{10.95 - 10}{10 \cdot 0.002}.\]
\[\Delta T = 475^\circ \text{C}.\]
Convert to Kelvin:
T = 475 + 273 = 748 K
Final Answer: $748 \, \text{K}$.
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).