Hi there,
If you build a non inverting amplifier with a fixed offset you should be able to get that kind of response. The amplifier would be built as follows (see attachment):
R1 connects from the inverting input to ground.
R2 connects from the output to the inverting input (feedback resistor).
R3 connects from the +5v supply to the inverting input (offset resistor).
The input connects to the non inverting input.
That's a simple configuration.
The defining equation for this circuit is:
Vo=((Vi*R2+Vi*R1)*R3+(Vi-5)*R1*R2)/(R1*R3)
where
Vo is the output voltage,
Vi is the input voltage,
5 is the reference voltage as discussed above.
Solving that equation for the two inputs of 2.700v and 2.710v and two outputs of 0.000v and 0.040v we get two equations in R2/R3 and R2/R1 from which we can find R2 and R3 after hand picking a value for R1. The general solutions are:
R2=R1*21/25, and
R3=R2*25/54 or R3=R1*7/18
Lets say we choose R1=10k, that makes R2=8.4k and R3=3.888k. These values give us 0v out at 2.700v in and for every additional 10mv we get 40mv additional output which means we get 40mv per deg C out.
Note however there will most likely be some input offset voltage which means the values of R2 and R3 may have to be adjusted slightly.
For example, for an op amp input offset of -1mv the values would change slightly to R2=8.408k and R3=3.894k, which may or may not affect the application significantly. For an input offset of +1mv the values would chance slightly to R2=8.392k and R3=3.884k, which again may or may not be significant to the application.
Note also that the +5v reference must be stable. It is possible to use a separate reference at a lower voltage if there will be a problem here, which would of course require modification of the two resistor values. For example, using a 4v reference would require changing that 5 in the defining equation to 4 and then rework the resistor values for R2 and R3.