Answer :
Answer:
[tex]47.94^{\circ}\text{C}[/tex]
Explanation:
[tex]m_1[/tex] = Initial mass of water = 100 g
c = Specific heat of water
[tex]\Delta T_1[/tex] = Temperature difference of the initial mass of water = [tex](24.8-x)^{\circ}\text{C}[/tex]
[tex]m_2[/tex] = Amount of water that is mixed = 75 g
[tex]\Delta T_2[/tex] = Temperature difference of the mixed mass of water = [tex](78.8-x)^{\circ}\text{C}[/tex]
[tex]x[/tex] = Equilibrium temperature
The heat balance of the system is
[tex]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}[/tex]
The final temperature of the mixed water is [tex]47.94^{\circ}\text{C}[/tex].