# Question 9b580

Jan 30, 2017

$q = - \text{9.4 J}$

#### Explanation:

The first thing to keep in mind here is that when ice cools it actually gives off heat to its surroundings.

This implies that $q$ will carry a negative sign since we associate a negative value of $q$ with heat being given off.

Now, the specific heat of ice tells you how much heat will be given off when the temperature of $\text{1 g}$ of ice decreases by ${1}^{\circ} \text{C}$ or by $\text{1 K}$.

More specifically, you now what

${c}_{\text{ice" = "2.087 J g"^(-1)"K"^(-1) = "2.087 J g"^(-1)""^@"C}}^{- 1}$

which means that when the temperature of $\text{1 g}$ of ice decreases by ${1}^{\circ} \text{C}$, $\text{2.087 J}$ of heat are being given off, i.e.

$q = - \text{2.087 J} \to$ corresponds to a decrease of ${1}^{\circ} \text{C}$ for $\text{1 g}$ of ice

You can start by calculating how much heat will be given off when $\text{0.12 g}$ of ice cools by ${1}^{\circ} \text{C}$. To do that, use the specific heat of ice as a conversion factor

0.12 color(red)(cancel(color(black)("g"))) * "2.087 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "0.25044 J"""^@"C"^(-1)

This means that $\text{0.25044 J}$ of heat are being given off when $\text{0.12 g}$ of ice is being cooled by ${1}^{\circ} \text{C}$. In your case, you must cool the sample by

DeltaT = |-66.0^@"C" - (28.6^@"C")|

$\Delta T = | - {37.4}^{\circ} \text{C"| = 37.4^@"C}$

which means that you will have

37.4 color(red)(cancel(color(black)(""^@"C"))) * overbrace("0.25044 J"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 0.12 g of ice")) = "9.4 J"#

This means that when $\text{0.12 g}$ of ice cools from $- {28.6}^{\circ} \text{C}$ to $- {66.0}^{\circ} \text{C}$, $\text{9.4 J}$ of heat are being given off, which is equivalent to saying that

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{q = - \text{9.4 J}}}}$

The answer is rounded to two sig figs.

So remember, when heat is being given off, $q$ carries a negative sign. When heat is being absorbed, $q$ carries a positive sign.