Can someone help me with the chemistry question below

Question

A piece of iron (Fe) weighing 40 g is heated to 800C and dropped into 100 g of water at 250C. (Specific heat of Fe is 0.446 J /g0C and Specific heat of H2O is 4.182 J /g0C) What is the temperature when thermal equilibrium has been reached?

Answer 1

Suppose temperature is T celsius

Then the iron loses (800-T) x 40 x 0.446 J heat

and the water gains (T - 250) x 100 x 4.182 J heat

These must be equal

17.84 (800-T) = 418.2 (T - 250)

14272 - 17.84 T = 418.2 T - 104550

118822 = 436.04 T

T = 272.50 celsius