Question

A500g chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating Styrofoam beaker containing of 1kgwater at room temperature $\mathbf{120}\mathbf{.}\mathbf{00}\mathbf{°}\mathit{C}$. After waiting and gently stirring for, you observe that the water’s temperature has reached a constant value of $\mathbf{22}\mathbf{°}\mathit{C}$.(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.

Short answer

the heat capacity of the metal is 215J/kgK

the water of eqaul weight will store more thermal energy

the Heat capacity of the metal will be smaller

Step-by-step solution

About Heat Capacity of substance

The heat capacity of a substance can be defined as the amount of heat required to change its temperature by one degree.

Thermodynamicsin its totality is concerned about heat. The meaning of heat today is energy in transit. Before the development of thermodynamic laws, the heat was considered as the measure of an invisible fluid, caloric, present in any matter

Determine the Specific heat of the metal

Givenm

${\mathrm{m}}_{\mathrm{metal}}=0.5\mathrm{kg},,{\mathrm{m}}_{\mathrm{w}}=1\mathrm{kg}$

For water ${\mathrm{T}}_1=20°\mathrm{C},,{\mathrm{T}}_2=22°\mathrm{C},,,{\mathrm{T}}_1=100°\mathrm{C}$

Solution

According to the given there is no change in phase so Q system: 0

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{w}}+{\mathrm{Q}}_{\mathrm{m}}+0 \\ {\mathrm{m}}_{\mathrm{w}}\times {\mathrm{C}}_{\mathrm{w}}\times {\mathrm{\delta T}}_{\mathrm{w}}=-{\mathrm{m}}_{\mathrm{m}}\times {\mathrm{C}}_{\mathrm{m}}\times \mathrm{Tm} \end{gathered}$$

By solving for Cm

$$\begin{gathered} \mathrm{Cm}=-\frac{{\mathrm{M}}_{\mathrm{w}}{\mathrm{C}}_{\mathrm{w}}{\mathrm{\delta T}}_{\mathrm{w}}}{{\mathrm{m}}_{\mathrm{m}}{\mathrm{\delta T}}_{\mathrm{m}}} \\ =-\frac{1\times 4190\mathrm{J}/\mathrm{kg}.\mathrm{K}\mathrm{x}{\left(22-20\right)}^0\mathrm{C}}{0.5\mathrm{kg}\mathrm{x}\left(22-100\right){\mathrm{C}}^0} \\ =215\mathrm{J}/\mathrm{kg}-\mathrm{K} \end{gathered}$$

Therefore the heat capacity of the metal is 215J/kgK

Determine which is more efficient for storing thermal energy

According to the result (a) We determine the specifc heat capacity of water higher than Cm of the metal, so water could absorb more heat and will store more Thermal energy-

Therefore the water of eqaul weight will store more thermal energy

Step 4:Determine the specific heat 

If the heat absorbed by Styrofoam not negligible the quantity heat of metal must be larger, so Cm in result would be smaller

Therefore the Heat capacity of the metal will be smaller