Answer:
The specific heat capacity of the unknown metal is 1.04 J/g°C.
Explanation:
Let the specific heat capacity of the unknown metal be [tex]x[/tex].
Given:
Mass of the metal is, [tex]m=55\ g[/tex]
Initial temperature of the metal is, [tex]T_i=90\ \°C[/tex]
Volume of water is, [tex]V=75\ ml[/tex]
Specific heat capacity of water is, [tex]c_w=4.186\ J/g\°C[/tex]
Initial temperature of water is, [tex]T_{wi}=25\ \°C[/tex]
Final temperature of the system is, [tex]T=35\ \°C[/tex]
We know that density of water is equal to 1 g/ml.
Mass is given as the product of density and volume.
Therefore, mass of water is given as:
[tex]m_w=1\times 75=75\ g[/tex]
Now, fall of temperature of the unknown metal is given as:
[tex]\Delta T_m=T_i-T=90-35=55\ \°C[/tex]
Rise of temperature of water is given as:
[tex]\Delta T_w=T-T_{wi}=35-25=10\ \°C[/tex]
Now, as per conservation of energy,
Heat lost by metal = Heat gained by water
⇒ [tex]mx\Delta T_m=m_wc_w\Delta T_w[/tex]
Plug in all the given values and solve for [tex]x[/tex]. This gives,
[tex]55\times x\times 55=75\times 4.186\times 10\\3025x=3139.5\\x=\frac{3139.5}{3025}=1.04\ J/g\°C[/tex]
Therefore, the specific heat capacity of the unknown metal is 1.04 J/g°C.
