To find the digit in the tens place, I need to look at the last two digits of $7^{2005}$. For example, $7^1 = 7, 7^2 = 49, 7^3 = 343, 7^4 = 2401, 7^5 = 16807, 7^6 = 117649, 7^7 = 823543, 7^8 = 5764801$, and so on. I can see that the last two digits of $7^1, 7^5, 7^9, \dots$ are 07, the last two digits of $7^2, 7^6, 7^{10}, \dots$ are 49, the last two digits of $7^3, 7^7, 7^{11}, \dots$ are 43, and the last two digits of $7^4, 7^8, 7^{12}, \dots$ are 01. So the last two digits depend on the remainder when the exponent is divided by 4. If the remainder is 0, the last two digits are 01. If the remainder is 1, the last two digits are 07. If the remainder is 2, the last two digits are 49. If the remainder is 3, the last two digits are 43. To find the remainder when 2005 is divided by 4, I can use long division or a calculator. I get 2005 = 4 \times 501 + 1, so the remainder is 1. This means the last two digits of $7^{2005}$ are the same as the last two digits of $7^1$, which are 07. Therefore, the digit in the tens place is 0. 
What is the digit in the tens place when $7^{2005}$ is expressed in decimal notation?