Two pumps are being filled by two different pumps. The first vat has 14 gallons is being filled at a rate of 20 gallons per second. The second vat has 35 gallons and is being filled at a rate of 16 gallons per second. After how many seconds will both vats have the same amount of liquid?

Let \( t \) be the time in seconds after both vats have started being filled.

For the first vat, which starts with 14 gallons and is being filled at a rate of 20 gallons per second, the amount of liquid in the first vat after \( t \) seconds can be expressed as:

\[
\text{Amount in first vat} = 14 + 20t
\]

For the second vat, which starts with 35 gallons and is being filled at a rate of 16 gallons per second, the amount of liquid in the second vat after \( t \) seconds can be expressed as:

\[
\text{Amount in second vat} = 35 + 16t
\]

We want to find \( t \) such that both vats have the same amount of liquid:

\[
14 + 20t = 35 + 16t
\]

To solve for \( t \), we first will simplify the equation by subtracting \( 16t \) from both sides:

\[
14 + 20t - 16t = 35
\]

This simplifies to:

\[
14 + 4t = 35
\]

Next, subtract 14 from both sides:

\[
4t = 21
\]

Now, divide both sides by 4:

\[
t = \frac{21}{4} = 5.25
\]

Thus, after \( t = 5.25 \) seconds, both vats will have the same amount of liquid.

To verify, we can calculate the amounts in both vats after 5.25 seconds:

1. First vat:

\[
14 + 20(5.25) = 14 + 105 = 119 \text{ gallons}
\]

2. Second vat:

\[
35 + 16(5.25) = 35 + 84 = 119 \text{ gallons}
\]

Since both vats have 119 gallons, the solution is confirmed.

The final answer is:

\[
\boxed{5.25}
\] seconds.