# Question 1c7a4

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 ${39.1}^{\circ} \text{C}$ to ${20.3}^{\circ} \text{C}$.

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 ${1}^{\circ} \text{C}$.

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

$q = m \cdot c \cdot \Delta T$, where

$m$ - the mass of the neon sample;
$c$ - the specific heat of neon;
$\Delta T$ - 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

$q = \textcolor{g r e e n}{- \text{250. J}}$