Answer:
[tex]T_2=17.8\°C[/tex]
Explanation:
Hello,
In this case, we can solve this problem by noticing that the heat lost by the warm water is gained by the ice in order to melt it:
[tex]Q_{water}=-Q_{ice}[/tex]
In such a way, the cooling of water corresponds to specific heat and the melting of ice to sensible heat and specific heat also that could be represented as follows:
[tex]m_{water}Cp_{water}(T_2-T_{water})=-m_{ice}\Delta H_{melting,ice}-m_{ice}\Cp_{ice}(T_2-T_{ice})[/tex]
Thus, specific heat of water is 4.18 J/g°C, heat of melting is 334 J/g and specific heat of ice is 2.04 J/g°C, thus, we can compute the final temperature as shown below:
[tex]m_{water}Cp_{water}(T_2-T_{water})+m_{ice}Cp_{ice}(T_2-T_{ice})=-m_{ice}\Delta H_{melting,ice}\\\\T_2=\frac{-m_{ice}\Delta H_{melting,ice}+m_{water}Cp_{water}T_{water}+m_{ice}Cp_{ice}T_{ice}}{m_{water}Cp_{water}+m_{ice}Cp_{ice}} \\\\T_2=\frac{-50.0*334+100*4.18*67+50.0*2.04*-20.0}{100*4.18+50.0*2.04} \\\\T_2=17.8\°C[/tex]
Best regards.
