Answer
Substituting the values of $C_{2}$ and $C_{3}$ in equation (1)
$998.2=C_{1}+(0.005) \times 20+(-0.005) \times 400$
$998.2=C_{1}+0.1-2$
$998.2=C_{1}=1.9$
$C_{1}=998.2+1.9$
$C_{1}=1000.1$
$\rho=1000.1+0.005 T-0.005 T^{2}$
At $T=42.1^{\circ} \mathrm{C}$
$\rho=C_{1}+C_{2} T-C_{2} T^{2}$
$\rho=1000.1+0.005 \times 42.1-0.005 \times 42.1^{2}$
$\rho=1000.1+0.2105-8.86205$
$\rho=991.45 \mathrm{~kg} / \mathrm{m}^{3}$
Work Step by Step
The empirical equation $\rho=C_{1}+C_{2} T+C_{3} T^{2}$
Where $C_{1}, C_{2}, C_{3}$ are constant
Considering density $=998.2$ at $20^{\circ} \mathrm{C}$
Considering density $=997.1$ at $25^{\circ} \mathrm{C}$
$997.1=C_{1}+C_{2} 25+C_{3} 625$
Subtracting equation (2) from equation (1)
$$
\begin{array}{ll}
1.1=-5 C_{2}-225 C_{3} & --(3)
\end{array}
$$
Considering density $=995.7$ at $30^{\circ} \mathrm{C}$
$995.7=C_{1}+C_{2} 30+C_{3} 900 \quad--(4)$
Considering density $=994.1$ at $35^{\circ} \mathrm{C}$
$\begin{array}{ll}994.1=C_{1}+C_{2} 35+C_{3} 1225 & --(5)\end{array}$
Subtracting equation (5) from equation (4)
$1.6=-5 C_{2}-325 C_{3} \quad--(6)$
Subtracting equation (3) from equation (6)
$$
0.5=-325 C_{3}-\left(-225 C_{3}\right)
$$
$0.5=-325 C_{3}+225 C_{3}$
$0.5=-100 C_{3}$
$C_{3}=-\frac{0.5}{100}$
$C_{3}=-0.005$
Substituting $C_{3}=-0.005$ in equation (6)
$1.6=-5 C_{2}-325(-0.005)$
$1.6=-5 C_{2}+1.625$
$5 C_{2}=1.625-1.6$
$C_{2}=\frac{0.025}{5}$
$C_{2}=0.005$
Substituting the values of $C_{2}$ and $C_{3}$ in equation (1)
$998.2=C_{1}+(0.005) \times 20+(-0.005) \times 400$
$998.2=C_{1}+0.1-2$
$998.2=C_{1}=1.9$
$C_{1}=998.2+1.9$
$C_{1}=1000.1$
$\rho=1000.1+0.005 T-0.005 T^{2}$
At $T=42.1^{\circ} \mathrm{C}$
$\rho=C_{1}+C_{2} T-C_{2} T^{2}$
$\rho=1000.1+0.005 \times 42.1-0.005 \times 42.1^{2}$
$\rho=1000.1+0.2105-8.86205$
$\rho=991.45 \mathrm{~kg} / \mathrm{m}^{3}$
