There are two different vats. The first vat has 20 gallons is being filled at a rate of 3 gallons per minute. The second vat has 108 gallons and is leaking fluid at the rate of 2 gallons per minute. After how many seconds will both vats have the same amount of liquid?

Let's denote the time in minutes as \( t \).

1. For the first vat:
- It starts with 20 gallons and is being filled at a rate of 3 gallons per minute.
- The amount of liquid in the first vat after \( t \) minutes is given by:
\[
\text{Amount in first vat} = 20 + 3t
\]

2. For the second vat:
- It starts with 108 gallons and is leaking at a rate of 2 gallons per minute.
- The amount of liquid in the second vat after \( t \) minutes is given by:
\[
\text{Amount in second vat} = 108 - 2t
\]

To find when both vats have the same amount of liquid, we set the two equations equal to each other:
\[
20 + 3t = 108 - 2t
\]

Now, let's solve for \( t \):

1. Add \( 2t \) to both sides:
\[
20 + 3t + 2t = 108
\]
This simplifies to:
\[
20 + 5t = 108
\]

2. Subtract 20 from both sides:
\[
5t = 108 - 20
\]
Which simplifies to:
\[
5t = 88
\]

3. Divide both sides by 5:
\[
t = \frac{88}{5} = 17.6 \text{ minutes}
\]

Now, to convert this into seconds:
\[
t = 17.6 \times 60 = 1056 \text{ seconds}
\]

Thus, both vats will have the same amount of liquid after **17.6 minutes** or **1056 seconds**.