\times 10^{-7} \, ^\circ \mathrm{C}^{-1} $.

Initial period of the pendulum, $ T_0 = 0.5 \, \mathrm{s} $.

Initial temperature, $ T_i = 20^\circ \mathrm{C} $.

Final temperature, $ T_f = 30^\circ \mathrm{C} $.

Determine the expansion in length due to temperature change:

$ \Delta l = l \times \alpha \times (T_f - T_i) $

Substituting the values:

$ \Delta l = l \times 9 \times 10^{-7} \times (30 - 20) $

Formulate the time period of the pendulum:

$ T = 2\pi \sqrt{\frac{l}{g}} $

Find the error in the time period:

$ \frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l} + \frac{1}{2} \frac{\Delta g}{g} $

Note: Since $ \Delta g = 0 $, this simplifies to:

$ \frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l} $

Substitute the changes into the equation:

$ \frac{\Delta T}{0.5} = \frac{1}{2} \left( \frac{9 \times 10^{-7} \times 10}{1} \right) $

Calculate $\Delta T$:

$ \Delta T = \frac{0.5 \times 9 \times 10^{-7} \times 10}{2} $

$ \Delta T = 2.25 \times 10^{-6} \, \mathrm{s} $

Thus, the clock loses $ 2.25 \times 10^{-6} $ seconds in each oscillation when the temperature increases to $ 30^\circ \mathrm{C} $.

Heat And Thermodynamics

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