Bernardino identifies unaltered dollar words

Python, 39 38 bytes

lambda s:sum(ord(i)-96for i in s)==100

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-1 byte thanks to @JonathanAllan

05AB1E, 8 bytes

Code:

Ç96-O4bQ

Uses the CP-1252 encoding. Try it online!

Explanation:

Ç          # Convert the string into a list of character codes
 96-       # Subtract 96 of each element
    O      # Take the sum
     4b    # Push 100 (4 in binary)
       Q   # Check if equal

Perl 6, 21 bytes

{100==[+] .ords X%32}

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Alternate:

{Ⅽ==[+] .ords X%32}

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Note that the Ⅽ is ROMAN NUMERAL ONE HUNDRED U+216D with a unival of … 100
Which takes 3 bytes to encode.

Expanded:

{  # bare block lambda with implicit parameter $_

  100     # is 100
  ==      # equal to
  [+]     # the sum of the following

    .ords # the ordinals of the input (implicit method call on $_)
    X[%]  # crossed using the modulus operator
    32    # with 32 to get 65..90 or 97..122 to become 1..26
}