Solution: The given curve is \[ C: z=t^{2}+\iota t,~0\le t \leq 2. \] So, we have \[ \mathrm{d} z = \mathrm{d} \left( t^{2}+\iota t \right) = \left( 2t + \iota \right) \mathrm{d}t. \]

Thus, the integral will be \begin{align*} \int_C \bar{z}\mathrm{d} z & = \int_0^2 \overline{\left( t^{2}+\iota t \right) } \left( 2t + \iota \right) \mathrm{d}t \\ & = \int _0^2 \left( t^2 -\iota t \right) (2t + \iota )\mathrm{d} t \\ & = \int _0^2 \left( 2t^3 + \iota t^2 -2\iota t^2 + t \right) \mathrm{d} t \\ & = \int _0^2 \left( 2t^3 - \iota t^2 + t \right) \mathrm{d} t \\ & = 2\int _0^2 t^3 \mathrm{d} t - \iota \int _0^2 t^2 \mathrm{d} t + \int _0^2 t\mathrm{d} t \\ & = 2 \left[ \frac{t^4}{4} \right]_0^2 - \iota \left[ \frac{t^3}{3} \right]_0^2 + \left[ \frac{t^2}{2} \right]_0^2 \\ & = 8 - \iota \frac{8}{3} + 2 \\ & = 10 - \frac{8}{3}\iota . \end{align*} Therefore, \[ \textcolor{blue}{\boxed{\int_C \bar{z}\mathrm{d} z = 10 - \frac{8}{3}\iota }} \]