iron with a mass of 25 grams is dropped into 100 grams of water. the change in temperature is 72. What is the temp change in the water?

Refer to: http://www.engineersedge.com/materials/specific_heat_capacity_of_metals_13259.htm
The specific heat of iron = 0.4605 J/(g°C)

It has been known that the specific heat of water = 4.186 J/g°C

Heat gained/lost = m c ΔT

Heat lost/gained by iron = Heat gained/lost by water
(25 g) × [0.4605 J/(g°C)] × (72°C) = (100 g) × [4.186 J/g°C] × ΔT

Temperature change in the water, ΔT = (72°C) × (0.4605/4.186) (25/100) = 1.98°C

You need something called the 'specific heat' ... it is in a table on your book.
Find the spec. heat of both water and iron.

Heat loss = heat gain

Q = mc(delta T) ... Q is heat loss or gain --- m = mass --- c = specific heat ----- delta T = change in temp.
Q = mc(delta T) <<< for iron
Q = 25(c iron)(72) <<< remember ... look up spec heat
Q = mc(delta T) <<< for water
Q = 100(c water)(delta T water) <<< remember ... look up spec heat

set the 2 equal and solve for delta T water
25(c iron)(72) = 100(c water)(delta T water) ... use a little algebra
... that is it