Answer :
Answer: [tex]0.543J/g^0C[/tex]
Explanation:
Volume of metal = volume of water displaced = (30.0 - 25.0) ml = 5.0 ml
Density of metal = 5.50 g/ml
Mass of metal =[tex]density\times volume =5.50\times 5.0=27.5g[/tex]
Volume of water = 25.0 ml
Density of metal = 1.0 g/ml
Mass of metal =[tex]density\times volume =1.0\times 25.0=25.0g[/tex]
[tex]heat_{absorbed}=heat_{released}[/tex]
As we know that,
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]-[m_1\times c_1\times (T_{final}-T_1)]=[m_2\times c_2\times (T_{final}-T_2)][/tex] .................(1)
where,
q = heat absorbed or released
[tex]m_1[/tex] = mass of metal = 27.5 g
[tex]m_2[/tex] = mass of water = 25.0 g
[tex]T_{final}[/tex] = final temperature = ?[tex]41.0^0C[/tex]
[tex]T_1[/tex] = temperature of metal = [tex]153^oC[/tex]
[tex]T_2[/tex] = temperature of water = [tex]25.0^oC[/tex]
[tex]c_1[/tex] = specific heat of lead = ?
[tex]c_2[/tex] = specific heat of water= [tex]4.184J/g^0C[/tex]
Now put all the given values in equation (1), we get
[tex]27.5g\times c_1\times (41.0-153)^0C=[25.0g\times 4.814J/g^0C\times (41.0-25.0)^0C][/tex]
[tex]c_1=0.543J/g^0C[/tex]
Thus the specific heat of the unknown metal sample is [tex]0.543J/g^0C[/tex]