The difference of temperature in a material can convert heat energy into electrical energy. To harvest the heat energy, the material should have:

### Related Formula V = S cdot Delta T$$V = S \cdot \Delta T$$ where, V$V$ = Thermoelectric voltage (Seebeck voltage) S$S$ = Seebeck coefficient Delta T$\Delta T$ = Temperature difference ### Core Logic To maximize the efficiency of a thermoelectric device harvesting heat energy, two conditions must be fulfilled: 1. **Low thermal conductivity**: This ensures that the temperature gradient (Delta T$\Delta T$) across the material is maintained and heat does not rapidly flow from the hot side to the cold side. 2. **High electrical conductivity**: This minimizes internal Joule heating losses (I^2R$I^2R$) when electrical current is drawn from the material. Therefore, the material should possess low thermal conductivity and high electrical conductivity. ### Pattern Recognition thermoelectric figure of merit is given by Z = fracS^2 sigmakappa$Z = \frac{S^2 \sigma}{\kappa}$, where sigma$\sigma$ is electrical conductivity and kappa$\kappa$ is thermal conductivity. To maximize Z$Z$, we inherently need high sigma$\sigma$ and low kappa$\kappa$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Current Electricity Class 11 Physics: Thermal Properties of Matter