Some notes to get you on your way:
Power in full sunlight at sea level $\approx 1kW/m^2$ - remember to adjust for relative angle of sun and solar panel.
Specific heat capacity of water: 4.186 J/g/K
Typical collector efficiency: somewhere in the range 50-90%.
Let's do a quick crude static calculation. It sounds like you're going to do a dynamic microsim - great stuff, go for it; if we can make a static estimate first, that'll give you an idea of how the numbers connect to each other, and give you a ballpark number to check the output of your simulation against.
Let's take the sun to be shining directly on the panel, along the normal to the plate, so the power hitting the plate is $1000 W/m^2$. Let's assume a 60% efficiency, so we're putting $600 W/m^2$ into the heat transfer fluid. Let's ignore any anti-freeze for the moment, and just assume our heat-transfer fluid is pure water, thus having a specific heat capacity of 4.186 J/g/K. Now, if you had 1 litre of water collecting heat per square metre, then you'd raise its temperature by
$$\frac{600 W}{1000 g \times 4.186 J /g/K} \approx 0.14 K/s$$
Now, here's a quick calculation for a whole day. Let's take a ball-park figure of 6 full sun-hours (that's 16 hours of summer daylight, derated by guesstimate to account for the variation of angle between panel and sun), and a 200-litre thermal storage tank; then our increase in temperature of the tank in a day, from a single square metre of collector would be, to the first order, be:
$$\Delta T_{tank} = \frac{0.14 K/s \times 3600 s/h \times 6 h}{200} \approx 15 K$$
So a $4m^2$ system would give you $\Delta T_{tank} \approx 60 K$ . Then that would get derated, based on system losses, including thermal losses from the collector plate and the thermal store.
Heat loss from the panel itself will (broadly) be proportional to the difference between the heat-transfer fluid, and the ambient air temperature - that gives the tapering you mentioned.
And you'll need a figure for the rate at which heat can be exchanged between the heat-transfer fluid that passes through the collector, and the thermal store.
(See also Appendix H of SAP 2009 which starts on pdf p73 - but bear in mind that that's a coarse static approximation of a solar thermal system in the UK, not a dynamic simulation - but it has some figures to get you started on collector efficiency and thermal losses)
Suppose you have a mass $M$ of oil at a temperature $T$ and specific heat capacity $C_{oil}$. You pump water at a temperature of $T_0$ and flow rate of $f$ kg/sec, and we'll assume as you say that the heating of the water by the oil is effectively instant. The heat absorbed by the water per second is:
$$ \frac{dU}{dt} = f \space \left(T - T_0 \right) \space C_w $$
where $C_w$ is the specific heat of water. The corresponding temperature drop per second in the oil is:
$$ \frac{dT}{dt} = \frac{1}{M \space C_{oil}} \frac {dU}{dt} = - \frac{f \space C_w}{M \space C_{oil}} \left(T - T_0 \right) $$
and one quick integration later we get:
$$ T - T_0 = T_{init} \space \text{exp} \left( - \frac{f \space C_w}{M \space C_{oil}} t \right) $$
where $T_{init}$ is the initial temperature in your tank i.e. 55C.
You give values for $f$ = 0.0833 kg/sec and $T_{init}$ = 55C. The specific heat of water, $C_w$ is 4.2 kJ/kg and oil is around 2 kJ/kg depending on the exact type of oil. However you don't give a value for M, so it's not possible to give an actual figure for the cooling rate. However if you plug in the mass of oil in your tank you can calculate the temperature as a function of time.
Best Answer
The convective heat transfer coefficient between the heater and its ambient will depend on the Nusselt number. The Nusselt number in turn is a function of the Reynolds number (forced convection) or the Rayleigh number (free convection), and the Prandtl number of the system. In case of free convection heat transfer the Rayleigh number is indeed a function of the delta_T between the heater and its surrounding, and hence it is expected that the heat transfer coefficient will also be dependent on the temperature difference. To get to know more on this scenario, you could refer to any standard heat transfer textbook like
http://web.mit.edu/lienhard/www/ahttv201.pdf