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A typical car has 16 L of liquid coolant circulating at a temperature of 95 ∘C through the engine's cooling system. Assume that, in this normal condition, the coolant completely fills the 2.0 −L volume of the aluminum radiator and the 14.0 −L internal cavities within the steel engine. When a car overheats, the radiator, engine, and coolant expand and a small reservoir connected to the radiator catches any resultant coolant overflow. Estimate how much coolant overflows to the reservoir if the system is heated from 95 ∘C to 105 ∘C. Model the radiator and engine as hollow shells of aluminum and steel, respectively. The coefficient of volume expansion for coolant is β=410⋅10−6/C∘

Respuesta :

Answer:

[tex]\Delta V=0.0592\ L[/tex]

Explanation:

Given:

  • Initial temperature of the coolant, [tex]T_f=95^{\circ}C[/tex]
  • final temperature of the coolant, [tex]T_f=105^{\circ}C[/tex]
  • total volume of the coolant, [tex]V_c=16\ L[/tex]
  • coefficient of volume expansion for coolant, [tex]\beta_c=410\times 10^{-6}\ ^{\circ}C^{-1}[/tex]
  • volume of Al radiator, [tex]V_a=2\ L[/tex]
  • volume of steel radiator,  [tex]V_s=14\ L[/tex]

We have:

coefficient of volume expansion for Aluminium, [tex]\beta_a=75\times 10^{-6}\ ^{\circ}C^{-1}[/tex]

coefficient of volume expansion for steel, [tex]\beta_a=35\times 10^{-6}\ ^{\circ}C^{-1}[/tex]

Now, change in volume of the coolant after temperature rises:

[tex]\Delta V_c=V_c.\beta_c.\Delta T[/tex]

[tex]\Delta V_c=16\times 410\times 10^{-6}\times (105-95)[/tex]

[tex]\Delta V_c=0.0656\ L[/tex]

Now, volumetric expansion in Aluminium radiant:

[tex]\Delta V_a=V_a.\beta_a.\Delta T[/tex]

[tex]\Delta V_a=2\times 75\times 10^{-6}\times (105-95)[/tex]

[tex]\Delta V_a=0.0015\ L[/tex]

Now, volumetric expansion in steel radiant:

[tex]\Delta V_s=V_s.\beta_s.\Delta T[/tex]

[tex]\Delta V_s=14\times 35\times 10^{-6}\times (105-95)[/tex]

[tex]\Delta V_s=0.0049\ L[/tex]

∴Total extra accommodation volume created after the expansion:

[tex]V_X=\Delta V_s+\Delta V_a[/tex]

[tex]V_X=0.0049+0.0015[/tex]

[tex]V_X=0.0064\ L[/tex]

Hence, the volume that will overflow into the small reservoir will be the volume of coolant that will be extra after the expanded accommodation in the radiator.

[tex]\Delta V=\Delta V_c-\Delta V_X[/tex]

[tex]\Delta V=0.0656-0.0064[/tex]

[tex]\Delta V=0.0592\ L[/tex]

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