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2 years ago

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A student obtained an unknown metal sample that weighed 65.3 g and at a temperature of 99.8oC, he placed it in a calorimeter con

taining 43.7 g of water at 25.7oC. At equilibrium the temperature of the water and metal was 34.5oC. Knowing the specific heat of the water to be 4.18 J/goC, what is the specific heat of the metal

1 answer:

lawyer [7]2 years ago

3 0

Answer:

0.377 J/gºC

Explanation:

From the question given above, the following data were obtained:

Mass of metal (Mₘ) = 65.3 g

Initial temperature of metal (Tₘ) = 99.8 °C

Mass of water (Mᵥᵥ) = 43.7 g

Initial temperature of water (Tᵥᵥ) = 25.7 °C

Equilibrium temperature (Tₑ) = 34.5 °C

Specific heat capacity of water (Cᵥᵥ) = 4.18 J/gºC

Specific heat capacity of metal (Cₘ) =?

The specific heat capacity of metal can be obtained as illustrated below:

Heat lost by metal = heat gained by water.

MₘCₘ(Tₘ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Cᵥᵥ)

65.3 × Cₘ (99.8 – 34.5) = 43.7 × 4.18 (34.5 – 25.7)

65.3Cₘ × 65.3 = 182.666 × 8.8

4264.09Cₘ = 1607.4608

Divide both side by 4264.09

Cₘ = 1607.4608 / 4264.09

Cₘ = 0.377 J/gºC

Therefore the specific heat capacity of the metal is 0.377 J/gºC

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Approximately $\rm -249.4\; kJ \cdot mol^{-1}$ .

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Note that hydrogen gas $\rm H_2\; (g)$ is the most stable allotrope of hydrogen. Since $\rm H_2$ is naturally a gas under standard conditions, the standard enthalpy of formation of $\rm H_2\; (g)$ would be equal to zero. That is:

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Look up the standard enthalpy of formation for the other species:

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(Source: CRC Handbook of Chemistry and Physics, 84th Edition (2004).)

$\displaystyle \Delta H^{\circ}_\text{reaction} = \sum \Delta H^{\circ}_f(\text{products}) - \sum \Delta H^{\circ}_f(\text{reactants})$ .

In other words, the standard enthalpy change of a reaction is equal to:

the sum of enthalpy change of all products, minus
the sum of enthalpy change of all reactants.

In this case,

$\begin{aligned} & \sum \Delta H^{\circ}_f(\text{products}) \\ =& \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + \Delta H^{\circ}_f(\mathrm{H_2O\;(g)})\end{aligned}$ .

$\begin{aligned} & \sum \Delta H^{\circ}_f(\text{reactants}) \\ =& \Delta H^{\circ}_f(\mathrm{CO\;(g)}) + 3\times \Delta H^{\circ}_f(\mathrm{H_2\;(g)})\end{aligned}$ .

Note that the number $3$ in front of $\Delta H^{\circ}_f(\mathrm{H_2\;(g)})$ corresponds to the coefficient of $\rm H_2$ in the chemical equation.

$\begin{aligned}&\Delta H^{\circ}_\text{reaction} \\ =& \sum \Delta H^{\circ}_f(\text{products}) - \sum \Delta H^{\circ}_f(\text{reactants})\\ =& \left(\Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + \Delta H^{\circ}_f(\mathrm{H_2O\;(g)})\right) \\ &- \left(\Delta H^{\circ}_f(\mathrm{CO\;(g)}) + 3\times \Delta H^{\circ}_f(\mathrm{H_2\;(g)})\right) \\ =& \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + (-74.6) - (3 \times 0 -110.5)\end{aligned}$ .

In other words,

$\begin{aligned} & \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + (-74.6) - (3 \times 0 -110.5) \\=& \Delta H^{\circ}_\text{reaction} = -213.5\; \rm kJ\cdot mol^{-1} \end{aligned}$ .

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