Answer:
[tex]m_{cold}=567kg\\\\m_{hot}=233kg[/tex]
Explanation:
Hello!
In this case, since equilibrium temperature problems involve the mass, specific heat and temperature change for the substances at different temperatures, we can write:
[tex]m_{cold}C_{cold}(T_{eq}-T_{cold})=-m_{hot}C_{hot}(T_{eq}-T_{hot})[/tex]
Thus, since we are talking about water and they both have the same specific heat, we can write:
[tex]m_{cold}(T_{eq}-T_{cold})=-m_{hot}(T_{eq}-T_{hot})[/tex]
Now, we plug in the temperatures to obtain:
[tex]m_{cold}(35-12)+m_{hot}(35-91)=0\\\\23m_{cold}-56m_{hot}=0[/tex]
Next, since the total volume of water is 800 L, since it has a density of 1kg/L, we infer the total mass is 800 kg; that is why we can write a 2x2 system of simultaneous equations:
[tex]\left \{ {{23m_{cold}-56m_{hot}=0} \atop {m_{cold}+m_{hot}=800}} \right.[/tex]
Thus, the masses of both cold and hot water turn out:
[tex]m_{cold}=567kg\\\\m_{hot}=233kg[/tex]
Best regards!
