Question

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

Answer 1

`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}`.