Chapter 6 - Sections 6.1-6.10 - Exercises - Problems by Topic - Page 289: 66

34.7 g

$q_{Fe}=-q_{water}$ $\implies m_{Fe}\times c_{Fe}\times\Delta T_{Fe}=-m_{water}\times c_{water}\times\Delta T_{water}$ $\implies m_{water}=-\frac{m_{Fe}\times c_{Fe}\times\Delta T_{Fe}}{c_{water}\times\Delta T_{water}}$ $=-\frac{(32.5\,g)(0.449\,J/g\cdot ^{\circ}C)(59.5^{\circ}C-22.7^{\circ}C)}{(4.18\,J/g\cdot ^{\circ}C)(59.5^{\circ}C-63.2^{\circ}C)}$ $=34.7\,g$