In an experiment on the specific heat of a metal, a 0.30 kg block of the metal at 120°Cis dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 100 $$\text{cm}^3$$​ ofwater at 27°C. The final temperature is 35°C. Calculate the specific heat of the metal?

Correct option is B

​​$$\begin{aligned}&\textbf{Given:} \\&\text{Mass of water } m_w = 0.025\, \text{kg} = 25\, \text{g}, \\&\text{Mass of metal } m = 0.30\, \text{kg} = 300\, \text{g}, \\&\text{Initial temperature of the metal } T_i = 120^\circ C, \\&\text{Final temperature } T_f = 35^\circ C, \\&\text{Specific heat of water } C_w = 4.186\, \text{J/g}^\circ \text{C}, \\&\text{Temperature change for the metal } \Delta T = 120 - 35 = 85^\circ C, \\&\text{Temperature change for the water } \Delta T_w = 35 - 27 = 8^\circ C.\end{aligned}$$

$$\begin{aligned}&\text{So, the mass of the water of } 100\, \text{cm}^3 \text{ volume } = 100\, \text{g} \\&\text{Specific heat of water, } C_w = 4.186\, \text{Jg}^{-1}\, \text{K}^{-1} \\&\Rightarrow M C_p T_w = (M + m) \times C_w \times (T_{wi} - T_f) \\&\text{Heat lost by metal = Heat gained by the water and calorimeter system} \\&\Rightarrow m C \Delta T_m = (M + m) \times C_w \times T_w \\&\Rightarrow 300 \times C \times (120 - 35) = (100 + 25) \times 4.186 \times (35 - 27) \\&\Rightarrow C \times 25500 = 10883.6 \\&\Rightarrow \boxed{C = 0.164\, \text{J}\, \text{g}^{-1}\, \text{K}^{-1}}\end{aligned}$$​​​

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