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12oz of water initially at 75oF is mixed with 20oz of water intiially at 140oF. What is the final temperature?

User CharlesAE
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1 Answer

4 votes

Answer:


115.625^(\circ)\text{F}

Step-by-step explanation:


m_1 = First mass of water = 12 oz


m_2 = Second mass of water = 20 oz


\Delta T_1 = Temperature difference of the solution with respect to the first mass of water =
(T-75)^(\circ)\text{F}


\Delta T_2 = Temperature difference of the solution with respect to the second mass of water =
(T-75)^(\circ)\text{F}

c = Specific heat of water

As heat gain and loss in the system is equal we have


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=(3700)/(32)\\\Rightarrow T=115.625^(\circ)\text{F}

The final temperature of the solution is
115.625^(\circ)\text{F}.

User Zerho
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