# Question fa6dc

Apr 5, 2015

To be able to solve this problem you need to know the values for the specific heat of lead and of water, which are known to be...

#### Explanation:

c_("lead") = "0.128 J/g "^@"C"

c_("water") = "4.18 J/g "^@"C"

The idea behind this problem is that the heat gained by the lead will be equal, in absolute value, to the heat lost by the water. This happens because the initial temperature of the lead weight is lower than the initial temperature of the water

q_("lead") = -q_("water"

The minus sign designates heat lost by the water. You can predict that the temperature of the lead weight will increase and the temperature of the water will decrease.

Mathematically, the above equation can be written as

${m}_{\text{lead") * c_("lead") * DeltaT_("lead") = m_("water") * c_("water") * DeltaT_("water}}$, where

${m}_{\text{lead}}$, ${m}_{\text{water}}$ - the masses of the lead weight and of the water, respectively;
$\Delta {T}_{\text{lead}}$, $\Delta {T}_{\text{water}}$ - the change in temperature for the lead weight and for the water, respectively.

Now, because the equilibrium temperature of the lead weight-water system is the same for both the lead weight and for the water, you can write

2.56cancel("g") * 0.128cancel("J")/(cancel("g") * ^@cancel("C")) * (color(red)(T_"eq") - 11.0^@cancel("C")) = -7.45cancel("g") * 4.18cancel("J")/(cancel("g") * ^@cancel("C")) * (color(red)(T_"eq") - 52.6^@cancel("C"))

To get rid of the minus sign, you can rewrite

$2.56 \cdot 0.128 \cdot \left(\textcolor{red}{{T}_{\text{eq") - 11.0) = 7.45 * 4.18 * (52.6 - color(red)(T_"eq}}}\right)$

Solving for $\textcolor{red}{{T}_{\text{eq}}}$ will give you

$0.32768 \cdot {T}_{\text{eq" - 3.60448 = -31.141 8 T_"eq}} + 1638.0166$

$30.81331 \cdot {T}_{\text{eq}} = 1634.4121$

${T}_{\text{eq}} = \frac{1634.41212}{30.81331}$

T_("eq") = 52.167^@"C"#

Rounded to three sig figs, the answer will be

${T}_{\text{eq") = color(green)("52.2"^@"C}}$

SIDE NOTE Notice how big the difference is between by how much the metal warms up and by how little the water cools down; this is due to the significant difference in specific heat values between lead and water.

The difference in mass contributes too, but not to the extent the difference in specific heat values does.

The video discusses how to solve a sample calorimetry calculation.

Video from: Noel Pauller