Answer : The initial temperature of water is, [tex]79.7^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 glass = [tex]0.840J/g^oC[/tex]
[tex]c_2[/tex] = specific heat of water = [tex]4.184J/g^oC[/tex]
[tex]m_1[/tex] = mass of glass = 32.50 g
[tex]m_2[/tex] = mass of water = 57 g
[tex]T_f[/tex] = final temperature of mixture = [tex]79.2^oC[/tex]
[tex]T_1[/tex] = initial temperature of glass = [tex]75^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = ?
Now put all the given values in the above formula, we get:
[tex]32.50g\times 0.840J/g^oC\times (79.2-75)^oC=-57g\times 4.184J/g^oC\times (79.2-T_2)^oC[/tex]
[tex]T_2=79.7^oC[/tex]
Therefore, the initial temperature of water is, [tex]79.7^oC[/tex]
