Answer to Question #217445 in Molecular Physics | Thermodynamics for fred

Question

Protons of mass m = 1.67 × 10−24 g and moving with a velocity v = 2×107 m/s strike a target of total mass M = 1g and heat capacity C = 0.334 cal/g/K; the proton beam carries a total current I = 4.8 μA. If 1/3 of the beam energy is converted into heat, calculate the temperature variation of the target in a time interval ∆t = 1 s.

Answer 1

With the information of the beam, we can find the number of protons if we know that the charge is equal to the charge of the proton times the number of protons:

$I=\frac{\Delta Q}{\Delta t}=\frac{N_pq_{proton}}{\Delta t} \implies N_p=\cfrac{I\Delta t}{q_{proton}}$

After that analysis we also have to calculate the energy of the beam, we use the equation for the kinetic energy of a single traveling proton and we find the total energy of the beam as the kinetic energy times the number of protons:

$E_{kinetic}=\frac{1}{2}m_{proton}v^2 \\ E_T=N_pE_{kinetic}=\cfrac{(I\Delta t)m_{proton}v^2}{2q_{proton}}$

Then, since only a third of the total energy is converted into heat we can relate that to the heat capacity and mass of the target, after substitution we find the temperature variation on the target as:

$Q=\frac{1}{3}E_T=\cfrac{(I\Delta t)m_{proton}v^2}{6q_{proton}}=M_{target}C_{target}\Delta T \\ \implies \Delta T=\cfrac{(I\Delta t)m_{proton}v^2}{6q_{proton}M_{target}C_{target}}$

$\Delta T=\cfrac{(4.8\times10^{-6}A)(1\,s)(1.67\times10^{-27}kg))(2\times10^7m/s)^2}{6(1.609\times10^{-19}C)(1\,g)(0.334\,cal/gK)(4.184\,J/cal)}$

After substitution this gives the temperature variation in Kelvin:

$\Delta T=\cfrac{(4.8)(1)(1.67))(2)^2}{6(1.609)(1)(0.334)(4.184)}K \\ \Delta T=2.3767\,K$

In conclusion, the temperature variation of the target in a time interval ∆t = 1 s is $\Delta T=2.3767\,K$.

Reference:

• Young, H. D., & Freedman, R. A. (2015). University Physics with Modern Physics and Mastering Physics (p. 1632). Academic Imports Sweden AB.