# What is the specific heat of the unknown metal sample?

## An irregular lump of an unknown metal has a measured density of 2.97 g/mL. The metal is heated to a temperature of 173 °C and placed in a graduated cylinder filled with 25.0 mL of water at 25.0 °C. After the system has reached thermal equilibrium, the volume in the cylinder is read at 34.0 mL, and the temperature is recorded as 40.6 °C. What is the specific heat of the unknown metal sample? Assume no heat is lost to the surroundings.

Feb 4, 2018

c=0.46 J/(g*°C)

#### Explanation:

Write down given information for both substances (mass, specific heat, change in temperature)

For Water :

$m = 25.0 g$

c = 4.181 J/(g*°C)

T_(i nitial) = 25°C

T_(fi nal) = 40.6°C

DeltaT = T_(fi nal)-T_(i nitial) = 40.6°C - 25°C = 15.6°C

For Metal :

$m = D \cdot V = 2.97 \frac{g}{m} L \cdot 9 m L = 26.73 g$

c = ?

T_(i nitial) = 173°C

T_(fi nal) = 40.6°C

DeltaT = T_(fi nal)-T_(i nitial) = 40.6°C - 173°C = - 132.4°C

Use the formula $Q = m c \Delta T$ to find the change in energy in water.

Q_w=(25.0 g)(4.181 J/(g*°C))(15.6°C)
${Q}_{w} = 1630.59 J$

This (${Q}_{w}$) is the amount of energy the water gained, so this means the metal lost the same amount($- {Q}_{m}$), according to the law of conservation of energy.

${Q}_{w} = - {Q}_{m}$

Rearrange the formula $Q = m c \Delta T$ to find $c$ of the metal.

$c = {Q}_{m} / \left(m \cdot \Delta T\right)$

c=(-1630.59J)/((26.73 g)(- 132.4°C))

c=0.46 J/(g*°C)

Feb 4, 2018

${C}_{m} \approx \text{0.457 J/g"^@ "C}$

The displacement of water gives the volume and thus mass of the metal:

$\text{34.0 - 25.0 mL" xx "2.97 g"/"mL" = "26.73 g}$

By conservation of energy,

${q}_{m} + {q}_{w} = 0$

where $m$ is metal and $w$ is water.

Thus

${q}_{m} = - {q}_{w}$

and

${m}_{m} {C}_{m} \Delta {T}_{m} = - {m}_{w} {C}_{w} \Delta {T}_{w}$

where $C$ is the constant-pressure specific heat capacity in $\text{J/g"^@ "C}$, ${m}_{i}$ the mass, and $\Delta T$ the change in temperature in $\text{^@ "C}$.

They both reach the same thermal equilibrium temperature, so

${m}_{m} {C}_{m} \left({T}_{f} - {T}_{m}\right) = - {m}_{w} {C}_{w} \left({T}_{f} - {T}_{w}\right)$

Therefore, assuming the specific heat capacity of water stays constant in the temperature range, and that its density is that at ${40.6}^{\circ} \text{C}$:

$\textcolor{b l u e}{{C}_{m}} = - \frac{{m}_{w} {C}_{w} \left({T}_{f} - {T}_{w}\right)}{{m}_{m} \left({T}_{f} - {T}_{m}\right)}$

= -("25.0 mL" xx "0.9919880 g"/"mL" cdot "4.184 J/g"^@ "C" cdot (40.6^@ "C" - 25.0^@ "C"))/("26.73 g" cdot (40.6^@ "C" - 173^@ "C"))

$= \textcolor{b l u e}{\text{0.457 J/g"^@ "C}}$