Answer:
[tex]T_{EQ}=42.1\°C[/tex]
Explanation:
Hello there!
In this case, since we are looking for an equilibrium temperature, we can evidence how the net heat flow is 0 as all the heat released by the hot brass is absorbed by the water:
[tex]Q_{brass}+Q_{water}=0[/tex]
Thus, by writing that equation in terms of mass, specific heat and temperature we obtain:
[tex]m_{brass}C_{brass}(T_{EQ}-T_{brass})+m_{water}C_{water}(T_{EQ}-T_{water})=0[/tex]
Thus, by applying the following algebra, it is possible to arrive to a clean expression to obtain the equilibrium temperature:
[tex]m_{brass}C_{brass}T_{EQ}+m_{water}C_{water}T_{EQ}=m_{water}C_{water}T_{water}+m_{brass}C_{brass}T_{brass}\\\\T_{EQ}(m_{brass}C_{brass}+m_{water}C_{water})=m_{water}C_{water}T_{water}+m_{brass}C_{brass}T_{brass}\\\\T_{EQ}=}\frac{m_{water}C_{water}T_{water}+m_{brass}C_{brass}T_{brass}}{m_{brass}C_{brass}+m_{water}C_{water}}[/tex]
Then, by plugging the masses, specific heats and initial temperatures in, we obtain:
[tex]T_{EQ}=\frac{130g*4.184\frac{J}{g*\°C} *35.0\°C+130g*0.380\frac{J}{g*\°C}*120.0\°C}{130g*4.184\frac{J}{g*\°C} +130g*0.380\frac{J}{g*\°C}}\\\\T_{EQ}=42.1\°C[/tex]
Best regards!
