Calorimetry Question Please

A sample of nickel is heated to 99.8 degrees Celsius. Then it is added to 150.0 grams of water at 23.5 degrees Celsius. The final temperature of the mixture is 45.2 degrees Celsius. The specific heat of water is 4.18 J/g * degrees Celsius and the specific heat of nickel is 0.444 J/g * degrees Celsius. What is the mass of nickel?

Any answers are appreciated... Explanations wouold be great too... Thank you in advance ^_^

MSDT=MSDT. Mass of water x specific heat of water x change in temp of water = MASS OF NICKEL x specific heat of nickel x change in temp of Nickel. 150g x 4.18J/g x 21.70C= M x .444 J/gx x 54.6oC. Move ur specific heat of Ni to the left as well as the change in temp. and u get.
((150)(4.18)(21.7))
((.444)(54.6)) = M (mass of nickel) Answer is 561g Ni. your welcome.

Well, we start with the essential equation, Q = m*c*(Tf - Ti) where Tf and Ti are final and initial temperatures respectively. You know that the amount of energy (heat) gained by the water has to equal the original amount of heat that the nickel had.
Energy gained by water = 150g * 4.18J/g * (45.2*C - 23.5*C) = 13605.9 J
This amount of energy had to be originally inside the X amount of nickel that you had.
But, you know that at the final temperature, EVERYTHING IS IN EQUILIBRIUM. This means that the temperature of the FINAL nickel is EQUAL to the temperature of the FINAL water.
So Ti = 99.8*C and Tf = 45.2*C for the nickel
Now you can just use the Q = m*c*(Tf-Ti) again, except this time you KNOW what Q is (solved for in the previous step with water.
Therefore, you get:
13605.9 J = m * 0.444 J/g * (99.8*C - 45.2*C)

m = 561 g of Nickel (3 significant figures because of 45.2 * C)