THERMAL expansionThermal expansion is an expansion of solids, liquids, or gases resulting from the increase in temperature or the temperature of that material.Expands> <shrinkIn general, any substance will expand when heated.The properties of a substance that terkit to increase and decrease the temperature of the thermal properties of substances called.A. Solids expansionThe theory of solid particles:Particles is compact and neatThe distance between the particles is very closeOf attraction (interaction) is very strong antarpartikelnyaFixed form
l = l0 ΔlDescription:l = l0 = length of first substancet1 = temperature at firstt2 = final temperature= temperature rise Δt = t2 - t1
The length:Δl ~ Δt → Δt = ΔlΔl ~ l0Solids Δl ~ what? Certainly always have a coefficient of linear expansion
α = number stating the length of 1 m when the temperature of the substance is dianikkan by 1 ° C or 1 Kα = ... / ° C or α = ... / Klt = long after raisedlo + lt = Δllo lo + lt = α Δtlt = u (1 + α Δt)
α = fraction of the length of objects with initial length per change in temperatureα = (Δl / l) / ΔtΔl / l = α ΔlΔl = l Δt
α = lim ┬ (ΔT → 0) 〖(Δl / l) / ΔT〗α = 1 / l lim ┬ (ΔT → 0) 〖ΔL / ΔT〗α = 1 / l dl / dT
l3
l1
γ = coefficient of volume expansionγ = (ΔV / V) / ΔT ↔ γ V ΔV = ΔTγ = 1 / V lim ┬ (ΔT → 0) 〖ΔV / ΔT〗γ = 1 / V dV / dT
Of V = l1 x l2 l3 xdV / dT = d / dT (x l1 l2 l3 x)
γ = 1 / V dV / dT = 1 / V d / dT (x l1 l2 l3 x)γ = 1 / V dl / dT + 1 l2 l3 / l1 l3 V dl2/dTγ = 1/l1 dl1/dT 1/l2 dl2/dT + + 1 / (l3) dl3/dTγ = α1 + α2 + α3α = α1 = α2 = α3
Solid when it expands, then it will happen strain (strain).S = Δl / l
Voltage / PressureF = WP = F / A
Young's modulusγ = stress / strain = (F / A) / (Δl / l)
Expansion of the Fluid
Composed of loose particles (distant)The interaction between particles (weak / less powerful)Follow containerIncompressible (not compressed)Expansion of the liquid (water) caused by the expansion of the container + γwadahExpansion by water + γcairPemuaiannya apparent expansion can be saidApparent coefficient of expansion: γs = γcair - γwadahVt = V0 (1 + ΔT. Γs) the container is assumedVt = V0 (1 + ΔT. Γcairan) does not expand
Anomaly of water is the event of depreciation of water when heated from the temperature 0o - 4 ° C, due to changes in the molecular bonds of water. At a temperature of 0o - 4 ° C, water has the greatest ρ.
Decrease in the density of matter: ρV0 = initial volumet = temperature riseVt = V0 (1 + γΔt)Vt> V0 Thermal Expansionρ0 = m0 / V0; ρt = m0 / Vtρ0> ρt Convectionρt <ρ0* M0 the mass of the substance that remains before and after reaction
Gas expansionExpansion of the gas can be done in several ways:Constant volumeAt constant volume, the change is the pressure (Pt).Pt = P0 (1 + Δt.γv)Here is a graph for a constant volume:
Constant pressure (for solids and liquids, thermal properties are not changed by pressure).At constant pressure, which changes the volume (Vt).Vt = V0 (1 + Δt.γp)Here is a graph for a constant pressure:
Based on the trial, apparently for all gas, γv = γp = 0.0036 = 1/273.Thus, Pt = P0 (1 + t/273); Vt = V0 (1 + t/273)
Volume and pressure changes.
Questions and DiscussionA 1000 m long steel bridge. When the temperature is raised from 0 to 30 ° C the increase in length is 0.33 m. What is the length of steel measuring 50 m if the temperature is raised from 0 to 60 ° C?Completion:Note: L1 = 1000 m, ΔT1 = 30 K, ΔL1 = 0.33 mL2 = 50 m, ΔT2 = 60 KAsked: ΔL2 = .......?Answer:ΔL_1 ΔT_1 L_1 = αΑ = ΔL1 / (l1 ΔT1)= (0.33 m) / (1000 m X 30 K)= 11 × 〖10〗 ^ (-6) / KΔL_ (2) = α. L_2. ΔT_2= 11 × 〖10〗 ^ (-6) / K x 50 m x 60 K= 0.033 mA 24 m length of copper, when heated the steel to be 24.02 m long. What is the temperature rise experienced by the steel. (Α = 17 × 〖10〗 ^ (-6) / K)Completion:Note: L_0 = 24 m, L1 = 24.02 m, α = 17 × 〖10〗 ^ (-6) / KAsked: ΔT = ......?Answer:ΔL = L - Lo= 24.02 m - 24 m= 0.02 m
ΔL = α. Lo. ΔT
ΔT = ΔL / (α Lo)= (0.02 m) / (17x〗 〖10 ^ (-6) .24 m)= (0.02 m) / (408 × 〖10〗 ^ (-6))= 50 K
A 1-liter glass vessel filled to the brim with mercury at 10 ° C. If the temperature rose to 50 ° C, how much mercury is spilled from the vessel?Completion:Note: ΔT = 40 KVkaca = 1 Literα = 9 x 10-6 / KWere: the amount of mercury is spilled?Answer:ΔVbejana = β V ΔT= 2 (9x10-6 / K). 1 L. 40 K= 10.8 x 10-4 MChanges in volume of mercury:ΔVair mercury = β V ΔT= 0.18 × 〖10〗 ^ (-3) L .1 .40 K= 72 x 〖10〗 ^ (-4) L
Thus, the amount of mercury spilled is72 x 10-4 - 10-4 = 10.8 x 61.2 x 10-2 Liter = 6.12 ml
l = l0 ΔlDescription:l = l0 = length of first substancet1 = temperature at firstt2 = final temperature= temperature rise Δt = t2 - t1
The length:Δl ~ Δt → Δt = ΔlΔl ~ l0Solids Δl ~ what? Certainly always have a coefficient of linear expansion
α = number stating the length of 1 m when the temperature of the substance is dianikkan by 1 ° C or 1 Kα = ... / ° C or α = ... / Klt = long after raisedlo + lt = Δllo lo + lt = α Δtlt = u (1 + α Δt)
α = fraction of the length of objects with initial length per change in temperatureα = (Δl / l) / ΔtΔl / l = α ΔlΔl = l Δt
α = lim ┬ (ΔT → 0) 〖(Δl / l) / ΔT〗α = 1 / l lim ┬ (ΔT → 0) 〖ΔL / ΔT〗α = 1 / l dl / dT
l3
l1
γ = coefficient of volume expansionγ = (ΔV / V) / ΔT ↔ γ V ΔV = ΔTγ = 1 / V lim ┬ (ΔT → 0) 〖ΔV / ΔT〗γ = 1 / V dV / dT
Of V = l1 x l2 l3 xdV / dT = d / dT (x l1 l2 l3 x)
γ = 1 / V dV / dT = 1 / V d / dT (x l1 l2 l3 x)γ = 1 / V dl / dT + 1 l2 l3 / l1 l3 V dl2/dTγ = 1/l1 dl1/dT 1/l2 dl2/dT + + 1 / (l3) dl3/dTγ = α1 + α2 + α3α = α1 = α2 = α3
Solid when it expands, then it will happen strain (strain).S = Δl / l
Voltage / PressureF = WP = F / A
Young's modulusγ = stress / strain = (F / A) / (Δl / l)
Expansion of the Fluid
Composed of loose particles (distant)The interaction between particles (weak / less powerful)Follow containerIncompressible (not compressed)Expansion of the liquid (water) caused by the expansion of the container + γwadahExpansion by water + γcairPemuaiannya apparent expansion can be saidApparent coefficient of expansion: γs = γcair - γwadahVt = V0 (1 + ΔT. Γs) the container is assumedVt = V0 (1 + ΔT. Γcairan) does not expand
Anomaly of water is the event of depreciation of water when heated from the temperature 0o - 4 ° C, due to changes in the molecular bonds of water. At a temperature of 0o - 4 ° C, water has the greatest ρ.
Decrease in the density of matter: ρV0 = initial volumet = temperature riseVt = V0 (1 + γΔt)Vt> V0 Thermal Expansionρ0 = m0 / V0; ρt = m0 / Vtρ0> ρt Convectionρt <ρ0* M0 the mass of the substance that remains before and after reaction
Gas expansionExpansion of the gas can be done in several ways:Constant volumeAt constant volume, the change is the pressure (Pt).Pt = P0 (1 + Δt.γv)Here is a graph for a constant volume:
Constant pressure (for solids and liquids, thermal properties are not changed by pressure).At constant pressure, which changes the volume (Vt).Vt = V0 (1 + Δt.γp)Here is a graph for a constant pressure:
Based on the trial, apparently for all gas, γv = γp = 0.0036 = 1/273.Thus, Pt = P0 (1 + t/273); Vt = V0 (1 + t/273)
Volume and pressure changes.
Questions and DiscussionA 1000 m long steel bridge. When the temperature is raised from 0 to 30 ° C the increase in length is 0.33 m. What is the length of steel measuring 50 m if the temperature is raised from 0 to 60 ° C?Completion:Note: L1 = 1000 m, ΔT1 = 30 K, ΔL1 = 0.33 mL2 = 50 m, ΔT2 = 60 KAsked: ΔL2 = .......?Answer:ΔL_1 ΔT_1 L_1 = αΑ = ΔL1 / (l1 ΔT1)= (0.33 m) / (1000 m X 30 K)= 11 × 〖10〗 ^ (-6) / KΔL_ (2) = α. L_2. ΔT_2= 11 × 〖10〗 ^ (-6) / K x 50 m x 60 K= 0.033 mA 24 m length of copper, when heated the steel to be 24.02 m long. What is the temperature rise experienced by the steel. (Α = 17 × 〖10〗 ^ (-6) / K)Completion:Note: L_0 = 24 m, L1 = 24.02 m, α = 17 × 〖10〗 ^ (-6) / KAsked: ΔT = ......?Answer:ΔL = L - Lo= 24.02 m - 24 m= 0.02 m
ΔL = α. Lo. ΔT
ΔT = ΔL / (α Lo)= (0.02 m) / (17x〗 〖10 ^ (-6) .24 m)= (0.02 m) / (408 × 〖10〗 ^ (-6))= 50 K
A 1-liter glass vessel filled to the brim with mercury at 10 ° C. If the temperature rose to 50 ° C, how much mercury is spilled from the vessel?Completion:Note: ΔT = 40 KVkaca = 1 Literα = 9 x 10-6 / KWere: the amount of mercury is spilled?Answer:ΔVbejana = β V ΔT= 2 (9x10-6 / K). 1 L. 40 K= 10.8 x 10-4 MChanges in volume of mercury:ΔVair mercury = β V ΔT= 0.18 × 〖10〗 ^ (-3) L .1 .40 K= 72 x 〖10〗 ^ (-4) L
Thus, the amount of mercury spilled is72 x 10-4 - 10-4 = 10.8 x 61.2 x 10-2 Liter = 6.12 ml
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