play phryzby with physics

Part of eval multi-llm branch, agent version multi-llm-servers · 2b11b97

Copper Block On Ice

solved Calorimetry

Q. A copper block of 2.5 kgmc=2.5kg is heated in a furnace to a temperature of 500 °CT1,c=500°C and then placed on a large ice blockT1,i=0°C. Specific heat of copper = 0.39 J g⁻¹ K⁻¹sc=0.39J g⁻¹ K⁻¹; heat of fusion of water = 335 J g⁻¹Li=335J g⁻¹. What is the maximum amount of ice that can meltT2,c=0°C, T2,i=0°C, Qc=−QL,i, in kg?

T = 0 °C500 °Ccopperm = 2.5 kgs = 390 J/kg·KQ0 °Cicem = ?L = 335000 J/kgQ
Inputs
mc
2.5 kg
T1,c
500 °C
sc
0.39 J g⁻¹ K⁻¹
Li
335 J g⁻¹
Outputs
ice meltmi?

How it's solved

Solved with Calorimetry

Given

  • mc=2.5kg
  • T1,c=500°C
  • sc=0.39J g⁻¹ K⁻¹
  • Li=335J g⁻¹
  • T1,i=0°C from “placed on a large ice block”
  • T2,c=0°C from “maximum amount of ice that can melt”
  • T2,i=0°C from “maximum amount of ice that can melt”
  • Qc=−QL,i from “maximum amount of ice that can melt”

Find mi.

Working
  1. T1,c=500°C=773.15K
  2. sc=0.39J g⁻¹ K⁻¹=390J kg⁻¹ K⁻¹
  3. Li=335J g⁻¹=335000J/kg
  4. T1,i=0°C=273.15K
  5. T2,c=0°C=273.15K
  6. T2,i=0°C=273.15K
  7. heat warms a body(copper)Qc=mc⋅sc⋅(T2,c-T1,c)
  8. Qc=2.5⋅390⋅(273.15-773.15)
  9. Qc=−487500J
  10. maximum amount of ice that can meltQc=−QL,i
  11. (−487500)=−QL,i
  12. QL,i=487500J
  13. heat changes a phase(ice)QL,i=mi⋅Li
  14. 487500=mi⋅335000
  15. mi=1.455kg
  16. mi≈1.5kgto 2 significant figures
Conclusion

The copper cools to 0 °C, giving up heat that melts ice; the maximum ice melted is when all the copper's heat goes into melting.

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
1question :copper_block_on_ice do2  kind :numerical3  describes "A copper block of %{copper_mass} is heated in a furnace to a temperature of %{copper_temp} and then placed on a large ice block. Specific heat of copper = %{copper_specific_heat}; heat of fusion of water = %{fusion}. %{ice_melt}, in kg?"4  given copper_mass: "2.5 kg"5  given copper_temp: "500 °C"6  given copper_specific_heat: "0.39 J g-1 K-1"7  given fusion: "335 J g-1"8  asks ice_melt: "What is the maximum amount of ice that can melt"9end
solution.rb
1solve :copper_block_on_ice, using: Calorimetry do2  objects :copper, :ice34  take copper_mass: copper.mass5  take copper_temp: copper.temperature_at_first6  take copper_specific_heat: copper.specific_heat_capacity7  take fusion: ice.latent_heat89  deduce_from "placed on a large ice block", ice.temperature_at_first => "0 °C"10  deduce_from "maximum amount of ice that can melt", copper.heat_to_warm => -(ice.heat_to_change_phase), copper.temperature_after => "0 °C", ice.temperature_after => "0 °C"11121314151617  1819  ice_melt = ice.mass20  answer ice_melt:, in: "kg", figures: 22122  conclusion "The copper cools to 0 °C, giving up heat that melts ice; the maximum ice melted is when all the copper's heat goes into melting."23end
playground.rb
1play_with :copper_block_on_ice do2  title "Copper Block On Ice"3end
The laws it drew on
quantitywrittenunitswithinfrom
massmkg0 … ∞Calorimetry
specific heat capacitysJ kg⁻¹ K⁻¹—Calorimetry
latent heatLJ/kg—Calorimetry
temperature at firstT1K—Calorimetry
temperature afterT2K—Calorimetry
heat to warmQJ—Calorimetry
heat to change phaseQLJ—Calorimetry
heat taken inQinJ—Calorimetry
heat to warm of groundQgJ—
mass of groundmgkg0 … ∞
specific heat capacity of groundsgJ kg⁻¹ K⁻¹—
temperature after of groundT2,gK—
temperature at first of groundT1,gK—
heat to warm of copperQcJ—
mass of coppermckg0 … ∞
specific heat capacity of copperscJ kg⁻¹ K⁻¹—
temperature after of copperT2,cK—
temperature at first of copperT1,cK—
heat to warm of iceQiJ—
mass of icemikg0 … ∞
specific heat capacity of icesiJ kg⁻¹ K⁻¹—
temperature after of iceT2,iK—
temperature at first of iceT1,iK—
heat to change phase of groundQL,gJ—
latent heat of groundLgJ/kg—
heat to change phase of copperQL,cJ—
latent heat of copperLcJ/kg—
heat to change phase of iceQL,iJ—
latent heat of iceLiJ/kg—
heat taken in of groundQin,gJ—
heat taken in of copperQin,cJ—
heat taken in of iceQin,iJ—
Qb=mb⋅sb⋅(T2,b-T1,b)
heat warms a body
QL,b=mb⋅Lb
heat changes a phase
Qin,b=Qb+QL,b
heat taken in warms and changes phase
Qg=mg⋅sg⋅(T2,g-T1,g)
heat warms a body(ground)
Qc=mc⋅sc⋅(T2,c-T1,c)
heat warms a body(copper)
Qi=mi⋅si⋅(T2,i-T1,i)
heat warms a body(ice)
QL,g=mg⋅Lg
heat changes a phase(ground)
QL,c=mc⋅Lc
heat changes a phase(copper)
QL,i=mi⋅Li
heat changes a phase(ice)
Qin,g=Qg+QL,g
heat taken in warms and changes phase(ground)
Qin,c=Qc+QL,c
heat taken in warms and changes phase(copper)
Qin,i=Qi+QL,i
heat taken in warms and changes phase(ice)
Qc=−QL,i
maximum amount of ice that can melt