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Answer :
Answer:
[tex]115.625^{\circ}\text{F}[/tex]
Explanation:
[tex]m_1[/tex] = First mass of water = 12 oz
[tex]m_2[/tex] = Second mass of water = 20 oz
[tex]\Delta T_1[/tex] = Temperature difference of the solution with respect to the first mass of water = [tex](T-75)^{\circ}\text{F}[/tex]
[tex]\Delta T_2[/tex] = Temperature difference of the solution with respect to the second mass of water = [tex](T-75)^{\circ}\text{F}[/tex]
c = Specific heat of water
As heat gain and loss in the system is equal we have
[tex]m_1c\Delta T_1=m_2c\Delta T_2\\\Rightarrow m_1\Delta T_1=m_2\Delta T_2\\\Rightarrow 12(T-75)=20(140-T)\\\Rightarrow 12T-900=2800-20T\\\Rightarrow 12T+20T=2800+900\\\Rightarrow 32T=3700\\\Rightarrow T=\dfrac{3700}{32}\\\Rightarrow T=115.625^{\circ}\text{F}[/tex]
The final temperature of the solution is [tex]115.625^{\circ}\text{F}[/tex].
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