Answer:
63.5F
Explanation:
Let x represent the temperature at the start of the experiment. Then the temperature at first check is (x -10.4F). The temperature at the second check is 1/3 that value, and is said to be 17.7F. This lets us write the equation ...
1/3(x -10.4F) = 17.7F
Multiplying by 3 gives ...
x -10.4F = 53.1F
x = 63.5F . . . . . . add 10.4F
The initial temperature of the liquid was 63.5F.
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Additional comment
Temperature is measured on an "interval scale." On such a scale, the value 0 is located arbitrarily. Concepts such as "1/3 of the temperature" have no meaning on such a scale. Here, we have to ignore that fact and assume the problem is about the numbers, not about any real temperature relationship in thermodynamic terms.