Question #1c7a4

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Stefan V.
Jul 15, 2015

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")#

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