I need help with a chemistry calculation to determine the heat of fusion for water.?

Question

Please show work.
(dHfus)(m ice)+(4.18J/g 0 C)(m ice)(T2) = -(4.18J/g 0 C)(m water)(T2-T1)
mass of water = 101.79g
Initial temperature = 23.5 C
Mass of Ice = 23.3g
Final temperature = 8.0 C

Answer 1

You're given all you need to solve the problem, just plug it in. It helps if you use all the units given, so you can make sure you don't make any mistakes. One note: ΔH is in J/g.

(ΔHfus)(m ice)+(4.18 J/g C)(m ice)(T2) = -(4.18J/g C)(m water)(T2-T1)

(ΔHfus J/g)(23.3g) + (4.18 J/g C)(23.3 g)(8.0 C) = -(4.18J/g C)(101.79g)(8 C - 23.5 C)

23.3(ΔHfus J/g) + 779.152 J = 6594.9741 J
23.3 g * (ΔHfus J/g) = 5815.8221 J
ΔHfus J/g = 249.606 J/g