# What was the heat lost from the system if "5.25 g" of water was cooled from 64.8^@ "C" to 5.5^@ "C"? C_P = "4.184 J/g"^@ "C"

Jan 16, 2018

Q_"lost"=ul?J

#### Explanation:

Set up the following processes to solve the problem with ease:

$G I V E N :$

$m = 5.25 g$, the mass of water

${C}_{p} = \frac{4.186 J}{g {\cdot}^{o} C}$, specific heat capacity of ${H}_{2} {O}_{l i q u i d}$; http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html

${T}_{f} = {5.5}^{o} C$, the initial temperature

${T}_{i} = {64.8}^{o} C$, the final temperature

$R E Q U I R E D :$

Q_"lost"=?

This is a negative value , as heat moves out from the system.

$S O L U T I O N :$

The heat flow is found based on the mass involved, its heat capacity, and the changes of temperature in the process; thus,

$Q = m {C}_{p} \Delta T$

where:

$\Delta T = {T}_{f} - {T}_{i}$

Now, plug in identified value to each variable in the formula as shown below. Make sure the units work out and the desired unit as required is attained :

$Q = m {C}_{p} \left({T}_{f} - {T}_{i}\right)$

$Q = 5.25 \cancel{g} \times \frac{4.186 J}{\cancel{g {\cdot}^{\circ} C}} \times \left(5.5 - 64.8\right) \cancel{{\cdot}^{\circ} C}$
Q=color(red)(-)ul ?J or

Q_"lost"=ul?J

Note:

The negative sign$\textcolor{red}{-}$ here indicates heat lost in the process.