Option 1 is correct

### Why use negative sign?

Lets understand this from a example,

Consider in the above diagram

at

\( T_{1}\)=400K, x=0m

\( T_{2}\)=100K, x=L=2m

k=1W/m-K

A=1\( m^{2}\)

We know that heat will transfer from higher temperature i.e 400K to lower temperature i.e 100K or from 0 to L.

Note: We are taking positive direction along the x-axis i.e from 0 to L. So heat transfer along this direction will be positive and vice versa.

Lets use the formula

\(\large Q=-kA\frac{\Delta T}{\Delta x}=-kA\frac{T_{2}-T_{1}}{2-0}=-kA\frac{(-300)}{2}=150J\)

So from the result we conclude that the rate of heat transfer is positive i.e from 0 to L direction.

In a similar fashion you can try considering the opposite way that the temperature at x=0m is 100K and temperature at x=2m is 400K.

\(\large Q=-kA\frac{\Delta T}{\Delta x}=-kA\frac{400-100}{2-0}=-kA\frac{(300)}{2}=-150J\)

We will find that the Q will come out to be negative since the heat transfer will take place from 2m to 0m i.e in negative x-direction and hence the direction of Q will be negative.

Note: The sign in the formula only denotes direction.