Question

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

Answer 1

Answer:

$115.625^{\circ}\text{F}$

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

$m_1c\Delta T_1=m_2c\Delta T_2\\\Rightarrow m_1\Delta T_1=m_2\Delta T_2\\\Rightarrow 12(T-75)=20(140-T)\\\Rightarrow 12T-900=2800-20T\\\Rightarrow 12T+20T=2800+900\\\Rightarrow 32T=3700\\\Rightarrow T=\dfrac{3700}{32}\\\Rightarrow T=115.625^{\circ}\text{F}$

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