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I'm going to guess that by "PT100" you mean a positive coefficient temperature sensor with a nominal resistance of 100 Ω at 0°C or 25°C.  Whatever it is you mean, it should be in your question.

This circuit has several independent parts, but that's hard to see at first glance due to how the schematic is drawn.  I'll describe what each part does separately.

<h2>Part 1</h2>

R2 and PT100 form a voltage divider.  R2 is fixed, but PT100 is not.  This part therefore produces a voltage that is a function of the PT100 resistance.  I'll assume (I shouldn't have to, you should state these things directly in your question) that the "120" below PT100 means that the nominal expected resistance of PT100 is 120 &Omega;.

The gain of the voltage divider is PT100 / (R2 + PT100).  At the nominal value of 120 &Omega; for PT100, that comes out to 0.0476.  That times the 2.5 V input yields 119 mV.

So, this part of the circuit produces a voltage that is a direct and known function of the PT100 resistance, nominally around 120 mV.

<h2>Part 2</h2>

R3, R5, and R4 both bias the opamp and provide negative feedback.  Note that although R3 is connected to R2, the two are really independent of each other because that connection point is held at a fixed 2.5 V.

The three resistors may look confusing, but the opamp is being run in classic non-inverting gain mode.  That means you expect a feedback resistor from the output back to the negative input, and another resistor from there to signal ground.

That's actually what you have, with the added wrinkle of the resistor from the negative input to signal ground being broken into two parts.  This can be more easily seen by realizing that V1, R3, and R5 form a Thevenin source.

The Thevenin voltage is V1 times the gain of the R3,R5 divider.  That comes out to 88 mV.  The resistance of the Thevenin source is R3//R5, which is 965 &Omega;.  I'll call that Rt.  Instead of R3 and R5, we can analyze the circuit as if there were 965 &Omega; to a 88 mV source instead.

We can now compute the overall amplifier gain, which is (R4 + Rt) / Rt = 35.4.  That gain is applied to variations from 88 mV.

<h2>Part 3</h2>

V2 provides the power to run the opamp.  We'll assume it's sufficient, and ignore it otherwise than noting the opamp can't swing past its supply rails, which in this case is 0 to 5 Volts.

<h2>Overall circuit</h2>

This circuit first converts the resistance signal of PT100 to a voltage, then amplifies that voltage around Vt.  I'll call the voltage across PT100 Vpt.  The amplifier output is therefore 35.4(Vpt - 88 mV) + 88 mV.

It would be useful to know what resistance range of PT100 the maximum possible output swing represents.  You can then look at the thermistor specs and find what temperature range the circuit is capable of measuring.  Note that real opamps can't drive all the way to their power rails, so the actual temperature range will be less.  I'll do the calculations for the full 0 V to 5 V output swing.  It's your job to find the real usable output swing and work back to the actual measurable temperature range.

The equation from before

&nbsp; &nbsp; Vout = 35.4(Vpt - 88 mV) + 88 mV

can be inverted to find Vpt as a function of Vout:

&nbsp; &nbsp; Vpt = (Vout - 88 mV)/35.4 + 88 mV

From this we find that at 0 V out Vpt is 85.5 mv, and at 5 V out Vpt is 227 mV.  I'll leave it as homework for you to find what resistances of PT100 those voltages represent, and in turn what temperatures.