Question

sample a
mass of each particle: 16 u
average particle speed: 1,500 m/s
sample b
mass of each particle: 28 u
average particle speed: 1,700 m/s
compare the average kinetic energies of the particles in each sample. which sample has the higher temperature?
sample a
sample b
neither, the samples have the same temperature

Explanation:

Step1: Recall kinetic - energy formula

The formula for the kinetic energy of a single particle is $K = \frac{1}{2}mv^{2}$, where $m$ is the mass and $v$ is the speed.

Step2: Calculate kinetic energy for sample A

For sample A, $m_A=16\ u$ and $v_A = 1500\ m/s$. First, convert the mass from atomic - mass units to kilograms. $1\ u=1.66\times10^{- 27}\ kg$, so $m_A = 16\times1.66\times10^{-27}\ kg$. Then $K_A=\frac{1}{2}m_Av_A^{2}=\frac{1}{2}\times(16\times1.66\times10^{-27})\times(1500)^{2}$.
\[

$$\begin{align*} K_A&=\frac{1}{2}\times16\times1.66\times10^{-27}\times2250000\\ &=8\times1.66\times10^{-27}\times2250000\\ &=8\times1.66\times2.25\times10^{-27 + 6}\\ &=8\times1.66\times2.25\times10^{-21}\\ &=29.88\times10^{-21}\ J \end{align*}$$

\]

Step3: Calculate kinetic energy for sample B

For sample B, $m_B = 28\ u$ and $v_B=1700\ m/s$. Convert the mass: $m_B = 28\times1.66\times10^{-27}\ kg$. Then $K_B=\frac{1}{2}m_Bv_B^{2}=\frac{1}{2}\times(28\times1.66\times10^{-27})\times(1700)^{2}$.
\[

$$\begin{align*} K_B&=\frac{1}{2}\times28\times1.66\times10^{-27}\times2890000\\ &=14\times1.66\times2.89\times10^{-27+6}\\ &=14\times1.66\times2.89\times10^{-21}\\ &=14\times4.8074\times10^{-21}\\ &=67.3036\times10^{-21}\ J \end{align*}$$

\]
Since the average kinetic energy of particles is directly proportional to the temperature of the substance (in the kinetic - theory of gases), and $K_B>K_A$, sample B has a higher temperature.

Answer:

sample B