In an experiment to determine the specific heat capacity of a solid, following observations were made:
Mass of calorimeter + stirrer is $x\,kg$
Mass of water is $y\,kg$
Initial temperature of water is ${t_1}\,{}^ \circ C$
Mass of the solid is $z\,kg$
Temperature of the solid is ${t_2}\,{}^ \circ C$
Temperature of mixture is $t\,{}^ \circ C$
Specific heat capacity of calorimeter and water are ${c_1}$ and ${c_2}$ respectively. Express the specific heat capacity $c$ of the solid in terms of the above data:
(A) $\dfrac{{\left( {x{c_1} + y{c_2}} \right)\left( {t - {t_1}} \right)}}{{z\left( {{t_2} - t} \right)}}$
(B) $\dfrac{{\left( {x{c_1} + y{c_2}} \right)\left( {t - {t_2}} \right)}}{{z\left( {{t_1} - t} \right)}}$
(C) $\dfrac{{\left( {x{c_1} + y{c_2}} \right)\left( {t + {t_2}} \right)}}{{z\left( {{t_1} + t} \right)}}$
(D) $\dfrac{{\left( {x{c_1} + y{c_2}} \right)\left( {t + {t_1}} \right)}}{{z\left( {{t_2} + t} \right)}}$

Hint: The specific heat of the solid is determined by equating the heat loss by the calorimeter and the water with the heat gained by the solid. The heat loss formula is used to make the heat loss equation of the calorimeter and water and heat gained formula is used to make the heat gained equation of the solid, then equating these two equations, the specific heat of the solid is determined.

Useful formula:
Heat loss or heat gain is given by,
$Q = mc\Delta T$
Where, $Q$ is the heat loss or gain, $m$ is the mass of the substance, $c$ is the specific heat and $\Delta T$ is the difference in the temperature.

Complete step by step solution:
Given that,
Mass of calorimeter + stirrer is $x\,kg$
Mass of water is $y\,kg$
Initial temperature of water is ${t_1}\,{}^ \circ C$
Mass of the solid is $z\,kg$
Temperature of the solid is ${t_2}\,{}^ \circ C$
Temperature of mixture is $t\,{}^ \circ C$
Specific heat capacity of calorimeter and water are ${c_1}$ and ${c_2}$ respectively.
Heat loss or heat gain is given by,
$Q = mc\Delta T\,.................\left( 1 \right)$
Now, the heat loss by the calorimeter is,
$Q = x{c_1}\left( {t - {t_1}} \right)\,...................\left( 2 \right)$
Now, the heat loss by the water is,
$Q = y{c_2}\left( {t - {t_1}} \right)\,...................\left( 3 \right)$
Now, the heat gained by the solid is,
$Q = z{c_3}\left( {{t_2} - t} \right)\,.................\left( 4 \right)$
The heat loss by the calorimeter and water is equal to the heat gained by the solid, then
$x{c_1}\left( {t - {t_1}} \right) + y{c_2}\left( {t - {t_1}} \right) = z{c_3}\left( {{t_2} - t} \right)$
By keeping the specific heat of the solid ${c_3}$ in one side, then
$\dfrac{{x{c_1}\left( {t - {t_1}} \right) + y{c_2}\left( {t - {t_1}} \right)}}{{z\left( {{t_2} - t} \right)}} = {c_3}$
By taking the term $\left( {t - {t_1}} \right)$ as a common, then
$\dfrac{{\left( {x{c_1} + y{c_2}} \right)\left( {t - {t_1}} \right)}}{{z\left( {{t_2} - t} \right)}} = {c_3}$
Thus, the above equation shows the specific heat of the solid.

Hence, the option (A) is correct.

Note: In equation (2) and equation (3), there is a heat loss so the temperature difference is mixed temperature to the initial temperature. But in equation (4), there is a heat gain so the temperature difference is the initial temperature of the solid to the mixed temperature.