Chapter 13, Problem 16P

To determine

The volume of the coolant that overflows from the radiator and the engine into the reservoir

Answer to Problem 16P

Solution:

The volume of the coolant that overflows from the radiator and the engine into the reservoir is $0.05628L$

Explanation of Solution

Given:

Initial volume of the radiator, $V_{Ri}=3.5L$

Initial volume of the engine, $V_{Ei}=10.5L$

Initial volume of the coolant (spread across the radiator and the engine),$V_{Ci}=14L$

Initial temperature, $T_i=93°C$

Final temperature,$T_f=105°C$

Coefficient of volume expansion for the coolant, ${\beta }_C=410\times {10}^{-6}\frac{1}{°C}$

Formula used:

$\Delta V=\beta V_i\Delta T$, where V_i is the initial volume and β is the coefficient of volume expansion

Calculation:

When the temperature increases, the volumes of the coolant, the radiator and the engine increase. Due to a higher rate of coefficient of volume, the expansion of the coolant as compared to aluminum, for the same temperature will be more; its increase in volume will be greater than that of the car parts. This will result in an overflow of the coolant, which will be collected by the reservoir.

The required total overflow can be obtained by finding the increase in volume of the radiator and the engine and subtracting from the increase in volume of the coolant.

The relation between the change in volume of a substance (ΔV)and the change in temperature (ΔT)is given by,

$\Delta V=\beta V_i\Delta T$, where V_i is the initial volume and Δ is the coefficient of volume expansion

Here, $\Delta T=T_f–T_i=(105°C-93°C)=12°C$

For aluminum, ${\beta }_A=75\times {10}^{-6}\frac{1}{°C}$

Therefore, increase in volume of the radiator, $\Delta V_R={\beta }_A{V}_{Ri}\Delta T$

 

and increase in volume of the engine, $\Delta V_E={\beta }_A{V}_{Ei}\Delta T$

 

Increase in volume of the coolant, $\Delta V_C={\beta }_C{V}_{Ci}\Delta T$

 

 

The volume of the overflowing coolant, $V_C’=\Delta V_C–\left(\Delta V_E+\Delta V_R\right)$

= ${\beta }_C{V}_{Ci}\Delta T–\left({\beta }_A{V}_{Ei}\Delta T+{\beta }_A{V}_{Ri}\Delta T\right)$

 

= $\left[{\beta }_C{V}_{Ci}-{\beta }_A\left(V_{Ei}+V_{Ri}\right)\right]\Delta T$

 

= $[((410\times {10}^{-6}\frac{1}{°C})\times \left(14L\right))–(75\times {10}^{-6}\frac{1}{°C})\left(10.5L+3.5L\right)]\left(12°C\right)$

$=0.05628L$