thanks for helping me out with the marathon question! it helped a lot !
im a bit confuse with this question though

a 150.0 g sample of metal at 75.0 degrees celcius is added to 150.0 g of H2O at 15.0 degrees celcius. the temperature of the water rises at 18.3 dgrees celcius. calculate the pacific heat capacity of the metal, assuming that all the heat lost by the metal is gain by the water

now heat lost by metal = heat gained by water
m1C1 delta t (metal)= m2C2 delta t (water)
C1 = m2 C2 delta t / m1 delta t= 150.0 g  x 4.18 x( 18.3-15)/ 150.0 g sample of metalX (118.3-75)
ignore the negative sign it just indicates that heat is liberated out.