Simple Binary to Decimal Converter
Your algorithm is ok, there are only a few minor hints for improvement:

You're using double-typed variables. Although in your case the computations will be exact up to fairly large numbers, it's generally safer to use int or long if you don't need the fractional part.

Two of your variable names begin with an upper-case letter (Binary and TempHold). That's unusual for Java code, as the vast majority of programs follows the naming conventions, so much that I find it irritating to read non-conforming code.

Your variable power has a name that confused me first. I'd call it exponent (in mathematics, "power" is the result of exponentiation). And I'd rename TempHold to digitChar and num to digitVal.

Math.pow() isn't really fast, so I recommend a different approach. With every iteration, the power grows by a factor of two, so I'd use something like power = power * 2; inside the loop.

For production code, you should separate input/output from computation, e.g. by putting the conversion part into a method of its own.

It's good style to close Scanners when they are finished (although that will eventually be done by Java's garbage collector if you don't do it explicitly).

So, my version of your program would be:

import java.util.Scanner;

public class BinaryToDecimal {

    public static void main(String[] args) {
        Scanner scan1 = new Scanner(System.in);
        System.out.println("Enter a binary number: ");
        String binary = scan1.next(); // 11011

        String decimal = toDecimal(binary);

        System.out.println(decimal + "(10)");
        scan1.close();
    }

    public static String toDecimal(String binary) {
        long power = 1;
        long sum = 0;

        for (int i = binary.length() - 1; i &gt;= 0; i--) {
            char digitChar = binary.charAt(i);
            int digitVal = Character.getNumericValue(digitChar);
            sum = sum + digitVal * power;
            power = power * 2;
        }
        return String.valueOf(sum);
    }
}