Question

Suppose you mix 100.0 g of water at 24.8 oC with 75.0 g of water at 78.8 oC. What will be the final temperature of the mixed water, in oC

Answer 1

Answer:

$47.94^{\circ}\text{C}$

Explanation:

$m_1$ = Initial mass of water = 100 g

c = Specific heat of water

$\Delta T_1$ = Temperature difference of the initial mass of water = $(24.8-x)^{\circ}\text{C}$

$m_2$ = Amount of water that is mixed = 75 g

$\Delta T_2$ = Temperature difference of the mixed mass of water = $(78.8-x)^{\circ}\text{C}$

$x$ = Equilibrium temperature

The heat balance of the system is

$m_1c\Delta T_1=m_2c\Delta T_2\\\Rightarrow 100(24.8-x)=75(x-78.8)\\\Rightarrow 2480-100x=75x-5910\\\Rightarrow -175x=-8390\\\Rightarrow x=\dfrac{8390}{175}\\\Rightarrow x=47.94^{\circ}\text{C}$

The final temperature of the mixed water is $47.94^{\circ}\text{C}$.