Dan places a bottle of water inside a cooler. As the water cools, its temperature $C(t)$ in degrees Celsius is given by the following function, where $t$ is the number of minutes since the bottle was placed in the cooler. \[ C(t)=7+17 e^{-0.038 t} \] Dan wants to drink the water when it reaches a temperature of 20 degrees Celsius. How many minutes should he leave it in the cooler? Round your answer to the nearest tenth, and do not round any intermediate computations.

Step 1 ：Given the function \(C(t)=7+17 e^{-0.038 t}\), where \(C(t)\) is the temperature of the water in degrees Celsius and \(t\) is the number of minutes since the bottle was placed in the cooler.

Step 2 ：We need to find the time \(t\) when the temperature \(C(t)\) reaches 20 degrees Celsius. So we set \(C(t) = 20\) and solve for \(t\).

Step 3 ：Subtract 7 from both sides of the equation to get \(17 e^{-0.038 t} = 13\).

Step 4 ：Divide both sides of the equation by 17 to get \(e^{-0.038 t} = \frac{13}{17}\).

Step 5 ：Take the natural logarithm of both sides to get \(-0.038 t = \ln\left(\frac{13}{17}\right)\).

Step 6 ：Finally, divide both sides by -0.038 to solve for \(t\).

Step 7 ：By calculating, we find that \(t\) is approximately 7.1.

Step 8 ：So, Dan should leave the bottle in the cooler for approximately \(\boxed{7.1}\) minutes.