Question

Suppose 61.0 g hot metal, which is initially at ${\mathbf{120}}^{\mathbf{0}}\mathit{C}$, is plunged into 100.0 g water that is initially at . The metal cools down and the water heats up until they reach a common temperature of $\mathbf{26}\mathbf{.}{\mathbf{39}}^{\mathbf{0}}\mathbf{C}$. Calculate the specific heat capacity of the metal, using$\mathbf{4}\mathbf{.}\mathbf{18}{\mathbf{JK}}^{\mathbf{-}\mathbf{1}}{\mathbf{g}}^{\mathbf{-}\mathbf{1}}$ as the specific heat capacity of the water.

Short answer

The specific heat capacity of the metal is .$0.468$

Step-by-step solution

Specific heat capacity

Specific heat capacity exhibits variation with temperature.Each state of matter possesses a unique value of specific heat capacity.

Calculating the specific heat capacity of the metal

The mass of the hot metal is 61.0 g.

The mass of water is 100.0 g.

The initial temperature of water is $20.{00}^{0}C$.

The initial temperature of the metal is $120.{00}^{0}C$.

The final temperature of the metal is $26.{39}^{0}C$

The specific heat capacity of the metal can be found as:

$$\begin{gathered} {\text{∆Q}}_{\text{sys}}{\text{=∆Q}}_{\text{metal}}{\text{+∆Q}}_{\text{water}}\text{= 0} \\ ∆{\text{Q}}_{\text{metal}}{\text{+∆Q}}_{\text{water}}{\text{=m}}_{\text{metal}}{\text{C}}_{\text{s,metal}}{\text{∆T}}_{\text{metal}}{\text{+m}}_{\text{water}}{\text{C}}_{\text{s,water}}{\text{∆T}}_{\text{water}}\text{= 0} \\ {\text{C}}_{\text{s,metal}}\text{=}\frac{{\text{m}}_{\text{water}}{\text{C}}_{\text{s,water}}{\text{∆T}}_{\text{water}}}{{\text{m}}_{\text{meta∆l}}{\text{T}}_{\text{metal}}} \end{gathered}$$

Where, ${\text{m}}_{\text{metal}}$and ${\text{m}}_{\text{water}}$denote the mass of metal and water respectively.

${\text{C}}_{\text{s,water}}$and ${\text{C}}_{\text{s,metal}}$denote the specific heat capacity of water and metal respectively.

${\text{∆T}}_{\text{water}}$and ${\text{∆T}}_{metal}$denote the temperature of the water and metal.

On further solving,

$$\begin{gathered} {\text{C}}_{\text{s,metal}}\text{=}\frac{{\text{m}}_{\text{water}}{\text{C}}_{\text{s,water}}{\text{×T}}_{\text{water}}}{{\text{m}}_{\text{metal}}{\text{×T}}_{\text{metal}}} \\ \text{=}\frac{{\text{100 g×4.18 JK}}^{\text{- 1}}{\text{g}}^{\text{- 1}}\text{×6.39 K}}{\text{61.0 g}\text{-×93.61 K}} \\ \text{=}\frac{{\text{2671.02JK}}^{\text{- 1}}{\text{g}}^{\text{- 1}}}{\text{5710.21}} \\ {\text{= 0.468JK}}^{\text{- 1}}{\text{g}}^{\text{- 1}} \end{gathered}$$

Hence, the specific heat capacity of the metal is ${\text{0.468 JK}}^{\text{- 1}}{\text{g}}^{\text{- 1}}$