Note these calculations are for: **Indium Doped Tin Oxide coated Glass (ITO) (100mm x 100mm), Sheet Resistance < 10 Ohm/sq, Size: 100mm x 100mm, Thickness: 1.1 mm**

Electric resistance **R** of ITO sheet 100x100mm (sheet resistance < 10 Ohm/sq) is 45 Ohm if electrodes are kept at the approximate distance of 100 mm.

Now assuming the condition, surrounding temperature is 25 °C and temperature has to be increased to 36 °C. So **temperature change ΔT is 11 °C which is 284.15 Kelvin.**

Electric Power **P**

$$P=\frac{V^2}{R}$$ here **V** is the Voltage applied.

Electric Energy **E**

$$[E=P*t=(\frac{V^2}{R})*t] -----(1)$$ here **t** is the time duration of Voltage applied.

Assuming temperature change of 284.15 K is to be achieved in 60 seconds (So time (t)= 60 sec). And also assuming heat loss in the surrounding is negligible or ideally Zero.

Specific heat **Cp** of **ITO (90%In _{2}O_{3}/10%SnO_{2})** is

Density **d** of **ITO (90%In _{2}O_{3}/10%SnO_{2})** is

$$Cp =\frac{E}{m* ΔT}$$ here **m** is the mass of ITO film

so, $$[E =Cp*m* ΔT] -----(2)$$

m= d*Volume of ITO Film ( Volume is around 1.85 * 10 ^{-3} cm^{3}, if we consider thickness of ITO film with sheet resistance < 10 Ohm/sq as 185nm and size of film to be 10cmx10cm )

Putting the value of d we get **m= 13.209 * 10 ^{-6} Kg**

now, combining equation (1) and (2) we get

$$Cp * m * ΔT = \frac{V^2}{R}*t$$

$$V=\sqrt{\frac{Cp * m * ΔT*R}{t}}$$

Putting all the values we get a **voltage = 0.963 V**. *So you need 0.963 V to increase temperature by 11 °C in 60 sec, if electrodes are kept 100 mm apart.*