Let \( t \) be the time in minutes after we start filling the vats.
For the first vat, which starts empty and fills at a rate of 6 gallons per minute, the amount of liquid in the vat after \( t \) minutes is given by:
\[ A(t) = 6t \]
For the second vat, which already contains 28 gallons and fills at a rate of 4 gallons per minute, the amount of liquid in the vat after \( t \) minutes is given by:
\[ B(t) = 28 + 4t \]
We want to find the time \( t \) when both vats have the same amount of liquid, so we set \( A(t) \) equal to \( B(t) \):
\[ 6t = 28 + 4t \]
Next, we will solve for \( t \):
- Subtract \( 4t \) from both sides:
\[ 6t - 4t = 28 \]
- Simplify the left side:
\[ 2t = 28 \]
- Divide both sides by 2:
\[ t = 14 \]
Now, we need to verify that this indeed gives both vats the same amount of liquid. We can substitute \( t = 14 \) back into the equations for the amounts of liquid in both vats.
- For the first vat:
\[ A(14) = 6 \times 14 = 84 \text{ gallons} \]
- For the second vat:
\[ B(14) = 28 + 4 \times 14 = 28 + 56 = 84 \text{ gallons} \]
Since both vats contain 84 gallons when \( t = 14 \), we conclude that the solution is verified. Therefore, the time when the two vats will have the same amount of liquid is
\[ \boxed{14} \text{ minutes}. \]