Question

Question #1c7a4

Answer 1

250. J are given off.

Explanation:

You actually need to know the specific heat of gaseous neon in order to be able to calculate the amount of heat released when that much neon is cooled from `39.1^@"C"` to `20.3^@"C"`.

The specific heat of neon is listed as

`c_"neon" = 1.03"J"/("g" ^@"C")`

A substance's specific heat tells you how much heat is needed/given off to increase/decrease the temperature of one gram of that substance by `1^@"C"`.

The equation that links added/released heat and change in temperature looks like this

`q = m * c * DeltaT`, where

`m` - the mass of the neon sample;
`c` - the specific heat of neon;
`DeltaT` - the change in temperature, defined as the final temperature minus the initial temperature.

Since the temperature of the sample is dropping, you should expect `q` to come out negative, which is an indicator that heat is being given off.

`q = 12.9cancel("g") * 1.03"J"/(cancel("g") * cancel(""^@"C")) * (20.3 - 39.1)cancel(""^@"C") = -"249.8 J"`

Rounded to three sig figs, the answer will be

`q = color(green)(-"250. J")`