Acopper calorimeter can with mass 0.100 kg contains 0.160 kg of water and 0.0180 kg of ice in thermal equilibrium at atmospheric pressure. if 0.750 kg of lead at 255c is dropped into the calorimeter can, what is the final temperature

First thing to do, you need the specific heats of the substances in the system and the heat of fusion of ice. From literature, they are as follows:

Copper = 0.385 kJ/kg.°C
Lead = 0.127 kJ/kg.°C
Water = 4.181 kJ/kg.°C
heat of fusion of ice = 333.6 kJ/kg 

Next, we let the final temperature be T in units of °C. 

The ice and water are in equilibrium thermally, so the temperature should be 0°C. By the law of conservation of energy,

Heat lost = Heat gained

Heat lost by lead = [ 0.75 (0.127 ) (255-T) ]kJ 

Heat from the melting ice = [(0.018 ) 333.6 ]kJ 
Heat gained from 0 to T = [(4.181) (T) (0.16 + 0.018)] kJ 
Heat to raise the calorimeter to T = 0.0385T]  kJ

Equating heat lost and heat gain, we can calculate for T. Therefore,

(0.75) (0.127) (255-T) = 6.0048 + 0.7442T + 0.0385T
T = 20.83 C 

Make them both to where they have the denominator of 18that makes 3/2 equal 27/18 and 1/9 equal 2/18so now you have  27/18x+5=2/18subtract 5 from the other sideso now you have 27/18x+2/18-5then subtract 2/18 from 27/18 to get 25/18x=-5then divide 25/18 by -5 and your answer will be  -5/18