Question

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∘

Answer 1

`\Delta V=0.0592\ L`

Step-by-step explanation:

Given:

• Initial temperature of the coolant,
  `T_f=95^(\circ)C`

• final temperature of the coolant,
  `T_f=105^(\circ)C`

• total volume of the coolant,
  `V_c=16\ L`

• coefficient of volume expansion for coolant,
  `\beta_c=410* 10^(-6)\ ^(\circ)C^(-1)`

• volume of Al radiator,
  `V_a=2\ L`

• volume of steel radiator,
  `V_s=14\ L`

We have:

coefficient of volume expansion for Aluminium,
`\beta_a=75* 10^(-6)\ ^(\circ)C^(-1)`

coefficient of volume expansion for steel,
`\beta_a=35* 10^(-6)\ ^(\circ)C^(-1)`

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

`\Delta V_c=V_c.\beta_c.\Delta T`

`\Delta V_c=16* 410* 10^(-6)* (105-95)`

`\Delta V_c=0.0656\ L`

Now, volumetric expansion in Aluminium radiant:

`\Delta V_a=V_a.\beta_a.\Delta T`

`\Delta V_a=2* 75* 10^(-6)* (105-95)`

`\Delta V_a=0.0015\ L`

Now, volumetric expansion in steel radiant:

`\Delta V_s=V_s.\beta_s.\Delta T`

`\Delta V_s=14* 35* 10^(-6)* (105-95)`

`\Delta V_s=0.0049\ L`

∴Total extra accommodation volume created after the expansion:

`V_X=\Delta V_s+\Delta V_a`

`V_X=0.0049+0.0015`

`V_X=0.0064\ L`

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.

`\Delta V=\Delta V_c-\Delta V_X`

`\Delta V=0.0656-0.0064`

`\Delta V=0.0592\ L`