Answer:
The final temperature of the system is 42.46°C.
Explanation:
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
[tex]q_1=-q_2[/tex]
[tex]m_1\times c\times (T_f-T_1)=-(m_2\times c\times (T_f-T_2))[/tex]
where,
c = specific heat of water= [tex]4.18J/g^oC[/tex]
[tex]m_1[/tex] = mass of water sample with 100 °C= 50.0 g
[tex]m_2[/tex] = mass of water sample with 13.7 °C= 100.0 g
[tex]T_f[/tex] = final temperature of system
[tex]T_1[/tex] = initial temperature of 50 g of water sample= [tex]100^oC[/tex]
[tex]T_2[/tex] = initial temperature of 100 g of water =[tex]13.7^oC[/tex]
Now put all the given values in the given formula, we get
[tex]50.0 g\times 4.184 J/g^oC\times (T_f-100^oC)=-(100 g\times 4.184 J/g^oC\times (T_f-13.7^oC))[/tex]
[tex]T_f=42.46^oC[/tex]
The final temperature of the system is 42.46°C.
