Answer : The unknown mass of water is, 200.3 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]
As,
[tex]c_1=c_2[/tex] = specific heat water
So,
[tex]m_1\times (T_f-T_1)=-m_2\times (T_f-T_2)[/tex]
where,
[tex]m_1[/tex] = mass of water = 150 g
[tex]m_2[/tex] = mass of unknown water = ?
[tex]T_f[/tex] = final temperature of mixture = [tex]67.3^oC[/tex]
[tex]T_1[/tex] = initial temperature of water = [tex]45^oC[/tex]
[tex]T_2[/tex] = initial temperature of unknown water = [tex]84^oC[/tex]
Now put all the given values in the above formula, we get
[tex]150g\times (67.3-45)^oC=-m_2\times (67.3-84)^oC[/tex]
[tex]m_2=200.3g[/tex]
Therefore, the unknown mass of water is, 200.3 grams.
