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Answer: The final temperature of both substances at thermal equilibrium is 301.0 K
Explanation:
As we know that,
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
.................(1)
where,
q = heat absorbed or released
= mass of gold = 31.5 g
= mass of water = 63.4 g
= final temperature = ?
= temperature of gold = 
= temperature of water = 
= specific heat of gold = 
= specific heat of water= 
Now put all the given values in equation (1), we get
![-31.5\times 0.129\times (T_{final}-342.4)=[63.4\times 4.184\times (T_{final}-300.4)]](https://tex.z-dn.net/?f=-31.5%5Ctimes%200.129%5Ctimes%20%28T_%7Bfinal%7D-342.4%29%3D%5B63.4%5Ctimes%204.184%5Ctimes%20%28T_%7Bfinal%7D-300.4%29%5D)
The final temperature of both substances at thermal equilibrium is 301.0 K
Answer:
55 g
Explanation:
First, we have to look for the solubility of KNO₃ at 60°C, considering that the solubility is the maximum amount of solute that can be dissolved in 100 grams of solvent, that is, the concentration of a saturated solution.
The solubility of KNO₃ at 60°C is 110.0 g of KNO₃ per 100 g of water. The mass of KNO₃ that must be dissolved in 50 g of water to make a saturated solution is:
50 g H₂O × (110.0 g KNO₃/100 g H₂O) = 55 g KNO₃
