Answer:
[tex]\Delta T_{w}=-0.715\°C[/tex]
Explanation:
Hello there!
In this case, according to the given information, it is possible to realize that the heat relationship between iron and water is given by:
[tex]-Q_{Fe}=Q_{w}[/tex]
Which in terms of mass, specific heat and temperature is:
[tex]-m_{Fe}C_{Fe}\Delta T_{Fe}=m_{w}C_{w}\Delta T_{w}[/tex]
Thus, by solving for the change in the temperature of water we will obtain:
[tex]\Delta T_{w}=\frac{-m_{Fe}C_{Fe}\Delta T_{Fe}}{m_{w}C_{w}} \\\\\Delta T_{w}=\frac{-200.0g*0.449\frac{J}{g\°C} 50.0\°C}{1500.0g4.184\frac{J}{g\°C}} \\\\\Delta T_{w}=-0.715\°C[/tex]
Which is negative because iron increases its temperature and therefore water decreases it.
Regards!
