2002
DOI: 10.1103/physrevb.66.224301
|View full text |Cite
|
Sign up to set email alerts
|

Colloidal metal particles as probes of nanoscale thermal transport in fluids

Abstract: We investigate suspensions of 3-10 nm diameter Au, Pt, and AuPd nanoparticles as probes of thermal transport in fluids and determine approximate values for the thermal conductance G of the particle/fluid interfaces. Subpicosecond ϭ770 nm optical pulses from a Ti:sapphire mode-locked laser are used to heat the particles and interrogate the decay of their temperature through time-resolved changes in optical absorption. The thermal decay of alkanethiol-terminated Au nanoparticles in toluene is partially obscured … Show more

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
1
1
1
1

Citation Types

15
276
1

Year Published

2006
2006
2022
2022

Publication Types

Select...
8
1

Relationship

0
9

Authors

Journals

citations
Cited by 285 publications
(298 citation statements)
references
References 31 publications
15
276
1
Order By: Relevance
“…Xue [74] found that the effects of the interfacial resistance, or the Kapitza resistance, between the nanotubes and the surrounding matrix can effectively reduce the thermal conductivity of the nanotubes by several orders of magnitude. Using the work of Huxtable et al [21], Wilson et al [75] and Xue [74] as justification, a phenomenological degradation of nanotube thermal conductivity by a factor of 20 is incorporated into the model. The data of Wang et al [23], Moisala et al [9], Hong & Tai [65], Biercuk et al [66], Cai & Song [24] and Xu et al [22] are selected to provide a broad spectrum of polymers and volume fractions for comparison.…”
Section: (Ii) Thermal Resultsmentioning
confidence: 99%
“…Xue [74] found that the effects of the interfacial resistance, or the Kapitza resistance, between the nanotubes and the surrounding matrix can effectively reduce the thermal conductivity of the nanotubes by several orders of magnitude. Using the work of Huxtable et al [21], Wilson et al [75] and Xue [74] as justification, a phenomenological degradation of nanotube thermal conductivity by a factor of 20 is incorporated into the model. The data of Wang et al [23], Moisala et al [9], Hong & Tai [65], Biercuk et al [66], Cai & Song [24] and Xu et al [22] are selected to provide a broad spectrum of polymers and volume fractions for comparison.…”
Section: (Ii) Thermal Resultsmentioning
confidence: 99%
“…(39). Due to the linearity of all the equations governing the heat release and diffusion in the system, the NP temperature at any time t after a series of N pulses at the repetition rate f is: Figure 8 plots the results of numerical simulations for two cases corresponding to two different NP radii.…”
Section: H Influence Of the Pulse Repetition Ratementioning
confidence: 99%
“…It has been demonstrated experimentally that an interface resistivity at the gold/water interface exists and can play a significant role in the heat release. [38][39][40][41] The interface resistivity can reach appreciable values when the liquid does not wet the solid. The wetting depends on the nature of the interface, and in particular a possible molecular coating.…”
Section: A Physical Systemmentioning
confidence: 99%
“…External heat flux in the system, Q ext , is nearly proportional to linear thermal driving force, with a heat transfer coefficient, h, as the proportionality constant: (9) In this case the energy balance in Eq 5 simplifies to: (10) Introducing a dimensionless driving force temperature, θ, scaled using the maximum system temperature, T max , (11) and a sample system time constant τ s (12) which are substituted into eq 10 and rearranged to yield (13) When laser irradiation ceases, Q I +Q o =0 and the system cools, reducing eq 13 to (14) Eq 14 may be solved using the initial condition θ = 1 at t = 0 to give (15) During laser irradiation, Q I +Q o is finite and system temperature rises to a maximum value when external heat flux given by eq 9 equals heat input via laser transduction given by eq 6: (16) Substituting eq 16 into eq 13 gives (17) Eq 17 can be solved using the initial condition θ = 0 at t = 0 to give (18) …”
Section: Heat Transfer Equation For Sample Cell and Contentsmentioning
confidence: 99%