Answer : The initial temperature of the water was [tex]43.2^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 thermometer = [tex]827J/kg.^oC[/tex]
[tex]c_2[/tex] = specific heat of water = [tex]4184J/kg.^oC[/tex]
[tex]m_1[/tex] = mass of thermometer = 30.3 g
[tex]m_2[/tex] = mass of water = 128 g
[tex]T_f[/tex] = final temperature of water and thermometer = [tex]41.8^oC[/tex]
[tex]T_1[/tex] = initial temperature of thermometer = [tex]11.9^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = ?
Now put all the given values in the above formula, we get:
[tex]30.3g\times 827J/kg.^oC\times (41.8-11.9)^oC=-128g\times 4184J/kg.^oC\times (41.8-T_2)^oC[/tex]
[tex]T_2=43.2^oC[/tex]
Therefore, the initial temperature of the water was [tex]43.2^oC[/tex]