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.
![q_1=-q_2](https://img.qammunity.org/2021/formulas/chemistry/college/ci1uvgegxwl3f5rpx3vscvsacjaiwja6yb.png)
![m_1* c_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)](https://img.qammunity.org/2021/formulas/chemistry/college/qyvqmycd0c0tziyigjg5lblqrbm4j5bbis.png)
As,
= specific heat water
So,
![m_1* (T_f-T_1)=-m_2* (T_f-T_2)](https://img.qammunity.org/2021/formulas/chemistry/high-school/jmgzfgkhy27j4qbxdz8k1xl3x786ghq6zh.png)
where,
= mass of water = 150 g
= mass of unknown water = ?
= final temperature of mixture =
![67.3^oC](https://img.qammunity.org/2021/formulas/chemistry/high-school/glorrpuutzwgbgh81diepsfn1wo7xuhqiq.png)
= initial temperature of water =
![45^oC](https://img.qammunity.org/2021/formulas/chemistry/high-school/ryqo6ppc212m15ze1cadv7sptggrgvjfin.png)
= initial temperature of unknown water =
![84^oC](https://img.qammunity.org/2021/formulas/chemistry/high-school/98i5g38glyy108kyei52gqn2wkgw3b58xb.png)
Now put all the given values in the above formula, we get
![150g* (67.3-45)^oC=-m_2* (67.3-84)^oC](https://img.qammunity.org/2021/formulas/chemistry/high-school/g5vqcxtp9d9wixnf2n68uqkdghligcsigf.png)
![m_2=200.3g](https://img.qammunity.org/2021/formulas/chemistry/high-school/dn6piofbznufkimt2a2ly9xvweukzzxgmk.png)
Therefore, the unknown mass of water is, 200.3 grams.