Dynamic response of temperature change when identical fluids flowing mix together 
As shown in the below figure , there is a mixing of a fuel in a system. the mass flow rates m1,m2 are different, the pipe diameters are all equal but are different in length.I need to get the temperature time response at the outlet after the mixing at length of 'a' . 
1) Is the temperature response linear or does it exponential ?
2) How would i apply the Newton law of cooling when the fluids are flowing ?
 A: 1. Mixing:
Assume liquid 1 has a specific heat capacity $c_{p,1}$ and liquid 2 $c_{p,1}$.
If the mass flows are respectively $\dot{m_1}$ and $\dot{m_2}$, at respective temperatures $T_1$ and $T_2$, then the temperature $T$ after perfect mixing (and assuming Enthalpy of mixing is zero) would be:
$$T=\frac{ c_{p,1}\dot{m_1}T_1+ c_{p,2}\dot{m_2}T_2}{c_{p,1}\dot{m_1}+c_{p,2}\dot{m_2}}$$
For identical fluids $c_{p,1}=c_{p,2}$ and the equation reduces to:
$$T=\frac{\dot{m_1}T_1+\dot{m_2}T_2}{\dot{m_1}+\dot{m_2}}$$
However, calculating the time $t$ required to achieve perfect mixing is much harder and depends on factors not provided. For example, mixing will be much faster is flow is turbulent, at high Reynolds ($Re$) number but $Re$ depends on flow speed, fluid viscosity and pipe diameter.
2. Cooling of mass flow in the pipe:
Let $\dot{m}$ be mass flow, $c_{p}$ heat capacity of the fluid, $T_0$ the surrounding’s temperature and $R$ the radius of the pipe.

Consider an infinitesimal element $\mathrm{d}x$. Due to heat loss it drops in temperature by $\mathrm{d}T$.
The heat loss acc. Newton is:
$$\frac{\mathrm{d}Q}{\mathrm{d}t}=h \, \mathrm{d}A(T-T_0)$$
where $\mathrm{d}A=2\pi R \, \mathrm{d}x$, so:
$$\frac{\mathrm{d}Q}{\mathrm{d}t}=2 \pi hR(T-T_0) \, \mathrm{d}x,$$
where $h$ is the coefficient of heat transfer.
This heat loss also causes a temperature drop $\mathrm{d}T$, so that:
$$\frac{\mathrm{d}Q}{\mathrm{d}t}=-c_p \, \mathrm{d}m\frac{\mathrm{d}T}{\mathrm{d}t}$$
(note the minus signs because $\mathrm{d}T<0$), where in fact:
$$\frac{\mathrm{d}m}{\mathrm{d}t}=\dot{m},$$
so that:
$$\frac{\mathrm{d}Q}{\mathrm{d}t}=-c_p \dot{m} \, \mathrm{d}T.$$
Equating both expressions for $\frac{\mathrm{d}Q}{\mathrm{d}t}$ yields:
$$2\pi hR(T-T_0) \, \mathrm{d}x=-c_p \dot{m} \, \mathrm{d}T.$$
This is a simple differential equation with separation of variables and after integration we obtain:
$$x=\frac{c_p \dot{m}}{2\pi h R} \ln\frac{T_1-T_0}{T-T_0},$$
where $T_1$ is the temperature at $x=0$.
If we set
$$\alpha=\frac{2\pi h R}{c_p \dot{m}},$$
then we find that
$$\frac{T-T_0}{T_1-T_0} = e^{-\alpha x}.$$
