A typical car has 18 L of liquid coolant circulating at a temperature of 95 C th

Question

A typical car has 18 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 3.5 L volume of the aluminum radiator and the 14.5 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 104 C. Model the radiator and engine as hollow shells of aluminum and steel, respectively. The coefficient of volume expansion for coolant is =410*10^6/C.

Explanation / Answer

We need to find the delta(V) or change in volume for both parts using:
dV = (V_o)(beta)(delta_T)
beta(aluminum) = 75 x 10^-6
beta(steel) = 35 x 10^-6

Aluminum: dV = (3.5L)(75 x 10^-6)(104 - 95) = .00236L
Steel: dV = (14.5L)(35 x 10^-6)(104-95) = .00456L

Now the NEW volume of each respective container is 3.50236L and 14.50456L, giving a new total volume of 18.00692L. We'll come back to this number.

(You could have done this next step first but whatever) Now we look at the volume expansion of the coolant:
dV = (18.0L)(410 x 10^-6)(104-95) = .06642L

What we want to know is how much spills over to the reservoir, which can be found by comparing the two new volumes we just calculated:

The coolant is now at a new volume of 18.06642L and the carrying capacity of the containers total 18.00692L. There is more coolant than space available! so just subtract it:
18.06642L - 18.00692L = .0595L, or59.5mL, or rounded to two sigfigs 60mL

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