Answer:
[tex]0.8696\ \text{J/g}^{\circ}\text{C}[/tex]
Explanation:
[tex]m_m[/tex] = Mass of metal = 19 g
[tex]c_m[/tex] = Specific heat of the metal
[tex]\Delta T_m[/tex] = Temperature difference of the metal = [tex]99-23=76^{\circ}\text{C}[/tex]
V = Volume of water = 150 mL = [tex]150\ \text{cm}^3[/tex]
[tex]\rho[/tex] = Density of water = [tex]1\ \text{g/cm}^3[/tex]
[tex]c_w[/tex] = Specific heat of the water = 4.186 J/g°C
[tex]\Delta T_w[/tex] = Temperature difference of the water = [tex]23-21=2^{\circ}\text{C}[/tex]
Mass of water
[tex]m_w= \rho V\\\Rightarrow m_w=1\times 150\\\Rightarrow m_w=150\ \text{g}[/tex]
Heat lost will be equal to the heat gained so we get
[tex]m_mc_m\Delta T_m=m_wc_w\Delta T_w\\\Rightarrow c_m=\dfrac{m_wc_w\Delta T_w}{m_m\Delta T_m}\\\Rightarrow c_m=\dfrac{150\times 4.186\times 2}{19\times 76}\\\Rightarrow c_m=0.8696\ \text{J/g}^{\circ}\text{C}[/tex]
The specific heat of the metal is [tex]0.8696\ \text{J/g}^{\circ}\text{C}[/tex].
