Answer
23.6 C
Work Step by Step
It is given that an ice cube at 0°C with a mass of 23.5 g is placed into 550.0g of water. Initially at 28.0 C. In an Insulated container.
Heat gained by the Ice = heat lost by the water= mc$\Delta t$
Enthalpy of fusion of water - 333.55 KJ/mol at 0 C
Specific heat of water - 4.1841/g"C
Heat gained by Ice = m (H_{Fusion}) + mc$\Delta t$ = (333.55 X23.5) + (24 X 4.184 X (T.0))
heat lost by water= 650 X4.184 X (24.T)
(333.55 X23.5) + [24 X 4.184 X (T-0)] = 650 X 4.184 X (24-T)
Solving the above equation for T,
T = 23.6 C
