A copper constant an thermocouple with it's cold junction at 0°c had an emf of 4.28mv when it's other hot junction was at 100°c. The emf 9.29mv when temperature of hot junction was 200°c. E=AQ+BQ^2?

To find the values of A and B in the equation E = AQ + BQ^2, we can use the given data points:

1. When the hot junction temperature is 100°C, the EMF is 4.28mV:
4.28 = A(100) + B(100)^2
4.28 = 100A + 10000B

2. When the hot junction temperature is 200°C, the EMF is 9.29mV:
9.29 = A(200) + B(200)^2
9.29 = 200A + 40000B

Now we have a system of two equations:

1. 4.28 = 100A + 10000B
2. 9.29 = 200A + 40000B

Solving these equations simultaneously will give us the values of A and B.

The given equation for the thermoelectric power of a thermocouple is:

```
E = AQ + BQ^2
```

where:

- E is the thermoelectric power (emf) in volts
- A and B are constants
- Q is the temperature difference between the hot and cold junctions in degrees Celsius

We have two sets of data:

- When the hot junction is at 100°C, E = 4.28 mV and Q = 100°C
- When the hot junction is at 200°C, E = 9.29 mV and Q = 200°C

We can use these data to solve for the constants A and B.

Substituting the first set of data into the equation, we get:

```
4.28 mV = A(0) + B(100)^2
```

```
4.28 mV = 10000 B
```

```
B = 0.000428 mV/°C^2
```

Substituting the second set of data into the equation, we get:

```
9.29 mV = A(0) + B(200)^2
```

```
9.29 mV = 40000 B
```

```
B = 0.000232 mV/°C^2
```

Therefore, the equation for the thermoelectric power of this thermocouple is:

```
E = 0.000428 Q + 0.000232 Q^2
```