Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of $\mathbf{3}\mathbf{.6}\mathbf{}\mathbf{\text{kg}}$and contains$\mathbf{14}\mathbf{}\mathbf{\text{kg}}$of water. A$\mathbf{1}\mathbf{.8}\mathbf{}\mathbf{\text{kg}}$piece of the metal initially at a temperature of$\mathbf{180}\mathbf{\text{°C}}$is dropped into the water. The container and water initially have a temperature of$\mathbf{16}\mathbf{.0}\mathbf{\text{°C}}$, and the final temperature of the entire (insulated) system is$\mathbf{18}\mathbf{.0}\mathbf{\text{°C}}$.

i) Mass of water$\left(M_w\right)=14\text{\hspace{0.17em}kg}$

ii) Mass of metal$\left(M_m\right)=1.8\text{\hspace{0.17em}kg}$

iii) Another mass$\left(M_c\right)\text{ofmetal}=3.6\text{\hspace{0.17em}kg}$

iv) Initial temperature$\left(T_{i1}\right)={180}^{\text{0}}\text{C}$

v) Initial temperature$\left(T_{i2}\right)={16}^{\text{0}}\text{C}$

vi) Final temperature of the entire system$\left(T_f\right)={18}^{\text{0}}\text{C}$

Using the concept of calorimetry, i.e., energy lost by a hot object is equal to the energy gained by the cold object when they reach equilibrium, we can write the equation for energy lost or gained by hot and cold objects respectively. We can use the formula for specific heat to write the equation for heat lost or heat gained in terms of mass, specific heat, and difference in the temperature. This equation can be solved to find the specific heat of the given material.

Formula:

The heat energy required by a body, …(i)

Where,$m$ = mass

$c$= specific heat capacity

$\Delta T$= change in temperature

$Q$= required heat energy

We can write the above formula as simplified for equilibrium using equation (i) as given:

$(M_w{c}_w+M_c{c}_m)(T_f-T_{i2})+M_m{c}_m(T_f-T_{i1})=0$

By solving forwe get the above formula as:

$\begin{array}{c}{c}_m=\frac{M_w{c}_w(T_{i2}-T_f)}{M_c(T_f-T_{i2})+M_m(T_f-T_{i1})}\\ =\frac{\left(14\text{\hspace{0.17em}kg}\right)\left(\frac{4.18\text{\hspace{0.17em}kJ}}{\text{kg.K}}\right)(16{}^{\text{0}}\text{C}-18{}^{\text{0}}\text{C})}{3.6\text{\hspace{0.17em}kg}({18}^{\text{0}}\text{C}-{16}^{\text{0}}\text{C})+1.8\text{kg}({18}^{\text{0}}\text{C}-{180}^{\text{0}}\text{C})}\\ =\frac{-117.04}{-284.4}\frac{\text{kJ}}{\text{kg.K}}\\ =0.411\frac{\text{kJ}}{\text{kg.K}}\end{array}$

Hence, the value of the specific heat is$0.411\frac{\text{kJ}}{\text{kg.K}}$