Answer:
[tex]T_F=47.3\°C[/tex]
Explanation:
Hello!
In this case, since we have hot and cold water, we infer that as hot water cools down, cool water heats up based on the first law of thermodynamics; thus, we can write:
[tex]Q_{hot}+Q_{cold}=0[/tex]
In such a way, we can write the expression in terms of mass, specific heat and temperature change:
[tex]m_{hot}C_{hot}(T_F-T_{hot})+m_{cold}C_{cold}(T_F-T_{cold})=0[/tex]
However, since they both have the same specific heat and the same mL are in g due to the 1.00-g/mL density, we obtain:
[tex]m_{hot}(T_F-T_{hot})+m_{cold}(T_F-T_{cold})=0\\\\T_F=\frac{m_{hot}T_{hot}+m_{cold}T_{cold}}{m_{hot}+m_{cold}}[/tex]
Now, we plug in to obtain:
[tex]T_F=\frac{140.0g*95.00\°C+300g*25.00\°C}{140.0g+300g}\\\\T_F=47.3\°C[/tex]
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