Question

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A 72.0-gram piece of metal at 96.0 °C is placed in 130.0 g of water in a calorimeter at 25.5 °C. The final temperature in the calorimeter is 31.0 °C. Determine the specific heat of the metal. Show your work by listing various steps, and explain how the law of conservation of energy applies to this situation.

Answer 1

Answer:

Approximately $0.639\; {\rm J \cdot g^{-1} \cdot K^{-1}}$.

Explanation:

Look up the specific heat of water:

$c(\text{water}) = 4.18\; {\rm J \cdot g^{-1} \cdot K^{-1}}$.

It is given that the mass of the water in this calorimeter is:

$m(\text{water}) = 130.0\; {\rm g}$,

The temperature of the water in this calorimeter has increased by:

$\Delta T(\text{water}) = (31.0 - 25.5)\; {\rm K} = 5.5\; {\rm K}$.

Thus, the energy that the water in this calorimeter absorbed would be:

$\begin{aligned}& Q(\text{water})\\ =\; & c(\text{water}) \, m(\text{water}) \, \Delta T(\text{water}) \end{aligned}$.

Let $c(\text{metal})$ denote the specific heat capacity of this piece of metal.

The mass of this piece of metal is:

$m(\text{metal})$.

The temperature of this piece of metal has been reduced by:

$\Delta T(\text{metal}) = (96.0 - 25.5)\; {\rm K} = 70.5\; {\rm K}$.

As this piece of metal cooled down, the energy it released would be:

$\begin{aligned}& Q(\text{metal})\\ =\; & c(\text{metal}) \, m(\text{metal}) \, \Delta T(\text{metal})\end{aligned}$.

Assuming that there is no exchange of energy between contents of this calorimeter and surroundings. By the law of conservation of energy, energy could neither be created nor destroyed.

Since energy could not be destroyed, the water in this calorimeter need to absorb the entirety of the energy that this piece of metal released. At the same, since no energy isn't created, the energy that the water absorbed would need to entirely come from this piece of metal. Therefore:

$Q(\text{metal}) = Q(\text{water})$.

$\begin{aligned}& c(\text{water})\, m(\text{water})\, \Delta T(\text{water}) \\&= c(\text{metal})\, m(\text{metal})\, \Delta T(\text{metal}) \end{aligned}$.

Rearrange and solve this equation for $c(\text{metal})$:

$\begin{aligned}& c(\text{metal}) \\=\; & \frac{c(\text{water}) \, m(\text{water})\, \Delta T(\text{water})}{m(\text{metal})\, \Delta T(\text{metal})} \\ =\; & \frac{4.18\; {\rm J \cdot g^{-1} \cdot K^{-1}} \times 130.0\; {\rm g} \times (31.0 - 25.5)\; {\rm K}}{72.0\; {\rm g} \times (96.0 - 25.5)\; {\rm K}} \\ \approx\; & 0.639\; {\rm J \cdot g^{-1}\cdot K^{-1}}\end{aligned}$.