Answer:
ΔT = 4.058 10²⁰

Step-by-step explanation:
In this experiment the system can be approximated as a tube with one end open and the other closed.
The open end is where the ultrasound emitter is and the closed end where the limit between the tissue and the bone is, the length of the tube is L = 0.55 cm = 5.5 10⁻³ m
a node is formed at the closed end and a belly at the open end, so the resonance has the form
λ = 4L 1st harmonic
λ = 4/3 L third harmonic
λ = 4/5 L fifth harmonic
λ = 4L / (2n + 1) n = 0, 1, 2, (2n + 1)
This wave is a standing wave therefore energy density remains in place
P = 1/2 ρ v (w S₀)²
angular velocity is related to frequency
w = 2π f
we substitute
E = P = 2π² ρ v f² S₀²
if this energy per unit area is transformed into heat
E = m c_e DT
let's use the concept of density
ρ = m / V
m = ρ V
if there are no losses in the system
½ π² ρ v f² S₀² = ρ V c_e ΔT
ΔT =
When analyzing this expression the temperature increase is
* quadratic at the frequency and maximum amplitude of the wave
* proportional to the speed of the wave in the tissue
* inversely proportional to tissue volume
we can approximate the volume of the tissue to the volume of a cylinder tube
V = π r² L
ΔT =
we calculate
ΔT = π 1450 (0.7 10⁶)² S₀² /( r² 5.5 10-3 c_e)
ΔT = 4.058 10²⁰
