Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

on the hot day in las vegas oil - trucker loaded v1 = 37*10^3 l of dies…

Question

on the hot day in las vegas oil - trucker loaded v1 = 3710^3 l of diesel fuel. he encountered cold weather on the way to payson, utah, where the temperature was δt = 23.0 k lower than in las vegas and where he delivered entire load. how many liters v2 did he deliver? the coefficient of volume expansion for diesel fuel γ = 9.510^(-4) c^(-1) and the coefficient of linear expansion for his steel truck tank α = 11*10^(-6) c^(-1).

Explanation:

Step1: Recall volume - expansion formula for fluid

The formula for the change in volume of a fluid due to temperature change is $\Delta V=\gamma V_1\Delta T$, where $V_1$ is the initial volume, $\gamma$ is the coefficient of volume expansion of the fluid, and $\Delta T$ is the change in temperature. The final volume of the fluid $V_{2,f}=V_1 + V_1\gamma\Delta T$.

Step2: Recall volume - expansion formula for the tank

The coefficient of volume expansion of a solid $\beta = 3\alpha$, where $\alpha$ is the coefficient of linear expansion of the solid. The change in volume of the tank $\Delta V_t=V_1\beta\Delta T=V_1\times3\alpha\Delta T$. The final volume of the tank $V_{2,t}=V_1 + V_1\times3\alpha\Delta T$.

Step3: Calculate the volume of diesel delivered

The volume of diesel delivered $V_2$ is given by considering the relative expansion of the diesel and the tank.
We know that $V_1 = 37\times10^{3}\text{ L}$, $\gamma=9.5\times 10^{-4}\text{ K}^{-1}$, $\alpha = 11\times10^{-6}\text{ K}^{-1}$, and $\Delta T=23.0\text{ K}$.
First, calculate the change in volume of the diesel $\Delta V_d=V_1\gamma\Delta T=(37\times 10^{3})\times(9.5\times 10^{-4})\times23.0$.
$\Delta V_d=37\times10^{3}\times9.5\times10^{-4}\times23.0 = 37\times9.5\times23\times10^{-1}=37\times218.5\times10^{-1}=808.45\text{ L}$.
Next, calculate the change in volume of the tank $\Delta V_t=V_1\times3\alpha\Delta T$.
$\Delta V_t=(37\times 10^{3})\times3\times(11\times10^{-6})\times23.0=37\times3\times11\times23\times10^{-3}=37\times759\times10^{-3}=28.083\text{ L}$.
Then $V_2=V_1+V_1\gamma\Delta T - V_1\times3\alpha\Delta T$.
$V_2=V_1+(V_1\gamma - V_1\times3\alpha)\Delta T$.
Substitute the values:
\[

$$\begin{align*} V_2&=37\times 10^{3}+(37\times 10^{3}\times9.5\times 10^{-4}-37\times 10^{3}\times3\times11\times 10^{-6})\times23.0\\ &=37\times 10^{3}+(37\times9.5\times10^{-1}-37\times33\times10^{-3})\times23.0\\ &=37\times 10^{3}+(35.15 - 1.221)\times23.0\\ &=37\times 10^{3}+33.929\times23.0\\ &=37\times 10^{3}+780.367\\ &=37780.367\text{ L} \end{align*}$$

\]

Answer:

$37780.367\text{ L}$