Answer
The beat frequency is $2.3~Hz$
Work Step by Step
At a temperature of $20.0^{\circ}C$, the speed of sound in the air is $343~m/s$. We can find the speed of sound in the air at a temperature of $25.0^{\circ}C$:
$v = 331+0.6~T$
$v = 331+0.6~(25.0^{\circ}C)$
$v = 346~m/s$
We can write an expression for the frequency of the pipe at the front:
$f_f = \frac{343~m/s}{\lambda}$
We can write an expression for the frequency of the pipe at the back:
$f_b = \frac{346~m/s}{\lambda}$
We can divide the two expressions to find $f_b$:
$\frac{f_b}{f_f} = \frac{\frac{346~m/s}{\lambda}}{\frac{343~m/s}{\lambda}}$
$f_b = \frac{346~m/s}{343~m/s}~f_f$
$f_b = \frac{346~m/s}{343~m/s}~(264.0~Hz)$
$f_b = 266.3~Hz$
Since the beat frequency is the difference between the two frequencies, the beat frequency is $2.3~Hz$
