Question

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.

Answer 1

`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*V = 2.97 g/mL * 9mL = 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=mcDeltaT` 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=mcDeltaT` to find `c` of the metal.

`c=Q_m/(m*DeltaT)`

`c=(-1630.59J)/((26.73 g)(- 132.4°C))`

`c=0.46 J/(g*°C)`

Answer 2

`C_m ~~ "0.457 J/g"^@ "C"`

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

`"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_mC_mDeltaT_m = -m_wC_wDeltaT_w`

where `C` is the constant-pressure specific heat capacity in `"J/g"^@ "C"`, `m_i` the mass, and `DeltaT` the change in temperature in `""^@ "C"`.

They both reach the same thermal equilibrium temperature, so

`m_mC_m(T_f - T_m) = -m_wC_w(T_f - T_w)`

Therefore, assuming the specific heat capacity of water stays constant in the temperature range, and that its density is that at `40.6^@ "C"`:

`color(blue)(C_m) = -(m_wC_w(T_f - T_w))/(m_m(T_f - T_m))`

`= -("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"))`

`= color(blue)("0.457 J/g"^@ "C")`