A 500.0-g iron bar at 212 degrees Celsius is placed in 2.0-L if water at 24.0 degrees Celsius. What will be the change in temperature of the water? Assume no heat is lost to the surroundings.

Use your formula Q = mc\Delta T

Since the no heat is lost, the magnitude (value) of both 'Q's are the same:

Q_{water} = Q_{iron}

m_{water}c_{water}\Delta T = m_{iron}c_{iron}\Delta T

(2\, kg)c_{water}(T_f - 24) = (0.5\, kg)c_{iron}(T_f-212)

You must have the specific heat capacity of water and iron. Replace c by their corresponding values and solve for Tf, the final temperature.

I hope it helped!:)