what is the fina equilibrium temperature when 40.0 grams of ice at 0°c is mixed with 5.00grams of stream at 120°c?

Heat change for changing temperature = m c ΔT
Heat change for changing physical state = m L

Specific heat capacity of steam = 2.08 J/(g K)
Latent heat of evaporation at 100°C = 2260 J/g
Specific heat capacity of water = 4.18 J/(g K)
Latent heat of fusion at 0°C = 334 J/g

Heat released :
for 5.00 g of 120°C steam to 100°C steam = 5.00 × 2.08 × (120 - 100) = 208 J
for 5.00 g of 100°C steam to 100°C water = 5.00 × 2260 = 11300 J
for 5.00 g of 100°C water to T°C water = 5.00 × 4.18 × (100 - T) = (2090 - 20.9T) J

Heat absorbed :
for 40.0 g of 0°C ice to 0°C water = 40.0 × 334 J/g = 13360 J
for 40.0 g of 0°C water to T°C water = 40.0 × 4.18 × (T - 0) = 167.2T J

Heat released = Heat absorbed
208 + 11300 + 2090 - 20.9T = 13360 + 167.2T
188.1T = 238
T = 1.3

The equilibrium temperature = 1.3°C

You need to look in your book for some information ...
specific heat (capacity) of water in all 3 forms (there will be 3 different values) one for liquid water, one for ice (solid) and one for steam (gas)
You also need to find the heat of fusion of water and the heat of vaporization of water.
==== without these you cannot do the problem.

The key to working the problem is to realize heat loss must = heat gain.

answer is approx 1.17 degrees C