Question

Most automobiles have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of copper and is filled to its 16.0-L capacity when at 10.0ºC. What volume of radiator fluid will overflow when the radiator and fluid reach their 95.0ºC operating temperature, given that the fluid volume coefficient of expansion is β = 400×10^{-6}/ ºC

Answer 1

After applying the formula for volumetric expansion and considering the initial volume and temperature change, it is found that 0.544 liters of coolant will overflow from the radiator when the temperature reaches 95.0°C.

Step-by-step explanation:

To calculate the volume of radiator fluid that will overflow due to thermal expansion, we need to use the formula for volumetric expansion, which is:

V = V₀ + V₀β(ΔT)

Where:

V is the final volume

• V0 is the initial volume
• β is the coefficient of volume expansion for the fluid
• ΔT is the change in temperature

Here, the initial volume V₀ is 16.0 L, β is 400×10⁻⁶/°C, and the temperature change ΔT (from 10.0°C to 95.0°C) is 85.0°C.

V0β(ΔT) = 16.0 L × (400×10⁻⁶ /°C × 85.0°C) = 0.544 L

Therefore, the final volume V is:

V = 16.0 L + 0.544 L = 16.544 L

The volume that will overflow is the difference between the final volume and the initial volume:

Volume overflowed = V - V₀ = 16.544 L - 16.0 L = 0.544 L

This means that 0.544 liters of coolant will overflow from the radiator when it reaches an operating temperature of 95.0°C.

Answer 2

There is a loss of fluid in the container of 0.475L

Step-by-step explanation:

To solve the problem it is necessary to take into account the concepts related to the change of voumen in a substance depending on the temperature.

The formula that describes this thermal expansion process is given by:

`\Delta V = \beta V_0 \Delta T`

Where,

`\Delta V =`Change in volume

`V_0 =`Initial Volume

`\Delta T =` Change in temperature

`\beta =` coefficient of volume expansion (Coefficient of copper and of the liquid for this case)

There are two types of materials in the container, liquid and copper, so we have to change the amount of Total Volume that would be subject to,

`\Delta V_T = \Delta V_l - \Delta V_c`

Where,

`\Delta V_l`= Change in the volume of liquid

`\Delta V_c`= Change in the volume of copper

Then replacing with the previous equation we have:

`\Delta V = \beta_l V_0 \Delta T- \beta_c V_0 \Delta T`

`\Delta V = (\beta_l-\beta_c)V_0\Delta T`

Our values are given as,

Thermal expansion coefficient for copper and the liquid to 20°C is

`\beta_c = 51*10^(-6)/\°C`

`\beta_l = 400*10^(-6)/\°C`

`V_0 = 16L`

`\Delta T = (95\°C-10\°C)`

Replacing we have that,

`\Delta V = (\beta_l-\beta_c)V_0\Delta T`

`\Delta V = (400*10^(-6)/\°C-51*10^(-6)/\°C)(16L)(95\°C-10\°C)`

`\Delta V = 0.475L`

Therefore there is a loss of fluid in the container of 0.475L