What will be the final temperature of 120.0 g of 20.0°C water when 100.0 g of 50.00°C iron nails are submerged in it?
(The specific heat of iron is 0.460 J/(g·°C).)

The final temperature will be 22.52°C.

Explanation:

Let the final temperature be T°C.

Heat gained by water = 120 × 4.186 × (T - 20)

                                    = 502.32 (T-20)

Heat lost by iron nails = 100 × 0.460 × (50 -T)

                                    = 46 (50-T)

As we know,

                         Heat lost = Heat gained

                   or, 46 (50-T) = 502.32 (T-20)

                  or, 2300 - 46T = 502.32T - 10046.4

         or, (2300 + 10046.4) = (502.32 + 46) T

                                    or, T =  

                                    ∴ T = 22.52°C

Hence the final temperature of the system is 22.52°C.