A 500 .0-g chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating Styrofoam beaker containing 1.00 kg of water at room temperature $\mathbf{20}\mathbf{°}\mathbf{C}$. After waiting and gently stirring for minutes, you observe that the water’s temperature has reached a constant value of $\mathbf{22}\mathbf{°}\mathbf{C}$. (a) Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat of the metal? (b) Which is more useful for storing thermal energy: this metal or an equal weight of water? Explain. (c) If the heat absorbed by the Styrofoam actually is not negligible, how would the specific heat you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain.

Step by step solution

01

Definition of metal

The term metal may be defined as the element having loses electrons.

02

Determine the specific heat of the metal, more useful for storing thermal energy, and the specific heat if the heat absorbed by the Styrofoam actually is not negligible.

Consider given data as below.

Mass of water, ${\mathrm{m}}_{\mathrm{w}}=1\mathrm{kg}$

Mass of metal, ${\mathrm{m}}_{\mathrm{m}}=500\mathrm{g}=0.5\mathrm{kg}$

Heat capacity, ${\mathrm{C}}_{\mathrm{w}}=4190\mathrm{J}/\mathrm{kg}°\mathrm{C}$

Temperature of water, ${\mathrm{T}}_{\mathrm{w}}=22°\mathrm{C}$

Room temperature, ${\mathrm{T}}_{\mathrm{R}}=20°\mathrm{C}$

Boiling temperature, $\mathrm{T}=100°\mathrm{C}$

There is no change in system. So,

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{system}}=0 \\ {\mathrm{Q}}_{\mathrm{w}}+{\mathrm{Q}}_{\mathrm{m}}=0 \\ {\mathrm{m}}_{\mathrm{w}}\times {\mathrm{C}}_{\mathrm{w}}\times {\mathrm{\delta T}}_{\mathrm{w}}=-{\mathrm{m}}_{\mathrm{m}}\times {\mathrm{C}}_{\mathrm{m}}\times {\mathrm{\delta T}}_{\mathrm{m}} \\ {\mathrm{C}}_{\mathrm{m}}=\frac{-{\mathrm{m}}_{\mathrm{w}}\times {\mathrm{C}}_{\mathrm{w}}\times {\mathrm{\delta T}}_{\mathrm{w}}}{{\mathrm{m}}_{\mathrm{m}}\times {\mathrm{\delta T}}_{\mathrm{m}}} \\ =\frac{-{\mathrm{m}}_{\mathrm{w}}\times {\mathrm{C}}_{\mathrm{w}}\times \left({\mathrm{T}}_{\mathrm{w}}-{\mathrm{T}}_{\mathrm{R}}\right)}{{\mathrm{m}}_{\mathrm{m}}\times \left({\mathrm{T}}_{\mathrm{w}}-\mathrm{T}\right)} \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} {\mathrm{C}}_{\mathrm{m}}=\frac{-1\times 4190\mathrm{J}/\mathrm{kg}.\mathrm{K}\times \left(22-20\right)°\mathrm{C}}{0.5\mathrm{kg}\times \left(22-100\right)°\mathrm{C}} \\ =215\mathrm{J}/\mathrm{kg}.\mathrm{K} \end{gathered}$$

Hence,the specific heat of the metal is 215 J/kg.K .

Hence, the conclusion is that ${\mathrm{HC}}_{\mathrm{w}}>{\mathrm{HC}}_{\mathrm{m}}$here stands on heat capacity, so water absorb more heat and will store more thermal energy. The heat absorbed by the Styrofoam actually is not negligible, the specific heat is smaller.

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