Given : Mass of solid $(m_1)=50 \mathrm{g}$ Initial temperature of solid $=150^{\circ} \mathrm{C}$ Mass of water $(m_2)=100 \mathrm{g}$ Intial temperature of mixture $=11^{\circ} \mathrm{C}$ Final temperature of mixture $=20^{\circ} \mathrm{C}$ Specific heat capacity of water $(c_2)=4.2 \mathrm{J} / \mathrm{g}^{\circ} \mathrm{C}$ By the principle of calorimetry, Heat lost by hot body = Heat gained by cold body. $\Rightarrow\; m_1 c_1 \Delta T_1\; = m_2 c_2 \Delta T_2$ $\Rightarrow 50 \times c_1 \times (150-20)\; = 100 \times 4.2 (20-11)$ $\Rightarrow 50 \times c_1 \times 130\; = 100 \times 4.2 \times 9$ $\Rightarrow\; c_1\; =\; \frac{100 \times 4.2 \times 9}{50 \times 30}$ $\;=\;0.58 \mathrm{Jg}^{-1}\,{}^{\circ}\mathrm{C}^{-1}$