Answer : The mass of the steel bar is, 35.2 grams.
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 steel = [tex]0.452J/g^oC[/tex]
[tex]c_2[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]
[tex]m_1[/tex] = mass of steel rod = ?
[tex]m_2[/tex] = mass of water = [tex]Density\times Volume=1.00g/mL\times 105mL=150g[/tex]
[tex]T_f[/tex] = final temperature of mixture = [tex]21.3^oC[/tex]
[tex]T_1[/tex] = initial temperature of steel = [tex]2.0^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = [tex]22.0^oC[/tex]
Now put all the given values in the above formula, we get
[tex]m_1\times (0.452J/g^oC)\times (21.3-2.0)^oC=-(105g)\times 4.18J/g^oC\times (21.3-22.0)^oC[/tex]
[tex]m_1=35.2g[/tex]
Therefore, the mass of the steel bar is, 35.2 grams.
