play phryzby with physics

A Hole in a Copper Sheet

solved ThermalExpansion

Q. A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cmL1=4.24cm at 27.0 °CT1=27.0°C. Coefficient of linear expansion of copper = 1.70 × 10⁻⁵ K⁻¹αl=1.70×10−5K⁻¹. What is the change in the diameter of the hole when the sheet is heated to 227 °CT2=227°C, in cm?

L₁ = 0.0424 mα = 1.7×10⁻⁵ /KL₂ΔL = ?27 °C227 °C
Inputs
L1
4.24 cm
T1
27 °C
αl
1.7e-05 K⁻¹
T2
227 °C
Outputs
changeΔl?

How it's solved

Solved with ThermalExpansion

Given

  • L1=4.24cm
  • T1=27.0°C
  • αl=1.70×10−5K⁻¹
  • T2=227°C

Find Δl.

Working
  1. L1=4.24cm=0.0424m
  2. T1=27.0°C=300.15K
  3. T2=227°C=500.15K
  4. linear expansion(hole)L2-L1=αl⋅L1⋅(T2-T1)
  5. L2-0.0424=1.7×10−5⋅0.0424⋅(500.15-300.15)
  6. L2=0.04254416m
  7. the change is the length after less the length at first(hole)Δl=L2-L1
  8. Δl=0.04254416-0.0424
  9. Δl=0.01442cm
  10. Δl=1.4416×10−4m=0.014416cm
  11. Δl≈0.0144cmto 3 significant figures
Conclusion

Heating swells the whole sheet like a photographic enlargement, so the hole grows too, just as a disc of copper filling it would.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb
1require_relative "../physics"23question :a_hole_in_a_copper_sheet do4  kind :numerical5  describes "A hole is drilled in a copper sheet. The diameter of the hole is %{diameter} at %{cool}. Coefficient of " \6            "linear expansion of copper = %{expansion}. %{change} when the sheet is heated to %{hot}, in cm?"7  given diameter: "4.24 cm"8  given cool: "27.0 °C"9  given expansion: "1.70e-5 K-1"10  given hot: "227 °C"11  asks change: "What is the change in the diameter of the hole"12end
solution.rb
1require_relative "../questions/a_hole_in_a_copper_sheet"2require_relative "../science/thermal_properties_of_matter/thermal_expansion"34solve :a_hole_in_a_copper_sheet, using: ThermalExpansion do5  objects :hole67  take diameter: hole.length_at_first8  take cool: hole.temperature_at_first9  take expansion: hole.coefficient_of_linear_expansion10  take hot: hole.temperature_after1112  change = hole.change_in_length13  answer change:, in: "cm", figures: 31415  conclusion "Heating swells the whole sheet like a photographic enlargement, so the hole grows too, just " \16             "as a disc of copper filling it would."17end
playground.rb
1require_relative "../solutions/a_hole_in_a_copper_sheet"23play_with :a_hole_in_a_copper_sheet do4  title "A Hole in a Copper Sheet"5end
The laws it drew on
quantitywrittenunitswithinfrom
length at firstL1m—ThermalExpansion
length afterL2m—ThermalExpansion
change in lengthΔlm—ThermalExpansion
temperature at firstT1K—ThermalExpansion
temperature afterT2K—ThermalExpansion
coefficient of linear expansionαlK⁻¹—ThermalExpansion
length after of groundL2,gm—
length at first of groundL1,gm—
coefficient of linear expansion of groundαl,gK⁻¹—
temperature after of groundT2,gK—
temperature at first of groundT1,gK—
length after of holeL2m—
length at first of holeL1m—
coefficient of linear expansion of holeαlK⁻¹—
temperature after of holeT2K—
temperature at first of holeT1K—
change in length of groundΔlgm—
change in length of holeΔlm—
L2,b-L1,b=αl,b⋅L1,b⋅(T2,b-T1,b)
linear expansion
Δlb=L2,b-L1,b
the change is the length after less the length at first
L2,g-L1,g=αl,g⋅L1,g⋅(T2,g-T1,g)
linear expansion(ground)
L2-L1=αl⋅L1⋅(T2-T1)
linear expansion(hole)
Δlg=L2,g-L1,g
the change is the length after less the length at first(ground)
Δl=L2-L1
the change is the length after less the length at first(hole)