Answer : The molar specific heat of Ni is, [tex]2.576J/mole^oC[/tex]
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
[tex]q_1=-q_2[/tex]
[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]
where,
[tex]c_1[/tex] = specific heat of Ni = ?
[tex]c_1[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]
[tex]m_1[/tex] = mass of Ni = 1000 g
[tex]m_2[/tex] = mass of water = 1 kg = 1000 g
[tex]T_f[/tex] = final temperature of water = [tex]26.3^oC[/tex]
[tex]T_1[/tex] = initial temperature of Ni = [tex]150^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = [tex]25^oC[/tex]
Now put all the given values in the above formula, we get
[tex]1000g\times c_1\times (26.3-150)^oC=-1000g\times 4.18J/g^oC\times (26.3-25)^oC[/tex]
[tex]c_1=0.0439J/g^oC[/tex]
Now we have to calculate the molar specific heat of Ni.
[tex]\text{Molar specific heat of Ni}=\text{Specific heat of Ni}\times \text{Molar mass of Ni}=(0.0439J/g^oC)\times (58.69g/mole)=2.576J/mole^oC[/tex]
Therefore, the molar specific heat of Ni is, [tex]2.576J/mole^oC[/tex]
