A sound wave travels at 1000 m/s in air at 0°C. What is its speed at 25°C?

The speed of sound in a medium is generally dependent on its temperature. For many materials, especially in introductory physics, the speed of sound is often approximated to be proportional to the square root of the absolute temperature (in Kelvin).

Given temperatures:
Initial temperature $T_1 = 0^\circ C$
Final temperature $T_2 = 25^\circ C$

Convert these to Kelvin:
$T_1 = 0 + 273.15 = 273.15 \text{ K}$
$T_2 = 25 + 273.15 = 298.15 \text{ K}$

The speed of sound ($v$) in a medium is often approximated to be proportional to the square root of its absolute temperature ($T$):

$$v \propto \sqrt{T}$$

This means we can write the ratio of speeds at two different temperatures as:

$$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$$

Given:
Initial speed $v_1 = 1000 \text{ m s}^{-1}$ at $T_1 = 273.15 \text{ K}$
We need to find $v_2$ at $T_2 = 298.15 \text{ K}$.

$$v_2 = v_1 \sqrt{\frac{T_2}{T_1}}$$ $$v_2 = 1000 \text{ m s}^{-1} \times \sqrt{\frac{298.15 \text{ K}}{273.15 \text{ K}}}$$

$$v_2 = 1000 \text{ m s}^{-1} \times \sqrt{1.091506}$$ $$v_2 = 1000 \text{ m s}^{-1} \times 1.04475$$ $$v_2 = 1044.75 \text{ m s}^{-1}$$

Rounding to a reasonable number of significant figures (e.g., 3 or 4, consistent with the given $1000 \text{ m s}^{-1}$):

$$v_2 \approx 1045 \text{ m s}^{-1}$$

The speed of sound at $25^\circ C$ is approximately $\boxed{\text{1045 m s}^{-1}}$.