Calculating specific heat of a metal

Mass of metal = 25.605g
Distilled Water Volume = 25mL
Distilled Water Temp = 25.3C
Temp of Metal = 100.5C
Temp of Mixture = 29.1C
(Specific Heat of water = 4.18J, in this given lab)

How do I figure out the specific heat of this metal?
Please Ive looked everywhere.
This is crucial. 10 Points for the first right answer.
Thank You Very Much

For this problem, you need to assume two things. First, that all of the heat released by the metal is absorbed by the water; and second, that the final temperatures of the water and of the metal are equal.

Since q = m * Cs * ΔT, where m is the mass, Cs is the specific heat, and ΔT is the temperature change, and the heat absorbed by the metal equals the heat released by the water, we can write the equation


m1 * Cs1 * ΔT1 = -1 * m2 * Cs2 * ΔT2

where m1 = mass of metal, Cs1 = specific heat of metal, ΔT1 = temperature change of metal, m2 = mass of water, Cs2 = specific heat of water, and ΔT = temperature change of water.

Since ΔT = final temp. - initial temp., make the substitutions ΔT1 = Tf1 - Ti1 and ΔT2 = Tf2 - Ti2

m1 * Cs1 * (Tf1 - Ti1) = -1 * m2 * Cs2 * (Tf2 - Ti2)

Now rearrange to solve for Cs1, since that' s the specific heat of the metal:

Cs1 = ( -1 * m2 * Cs2 * (Tf2 - Ti2) ) / ( m1 * (Tf1 - Ti1) )

Now plug in the following:
m2 = mass of water = 25 g (since 1 mL H2O = 1 g)
Cs2 = specific heat of water = 4.18 J/ g °C
Tf2 = Final temp. of water = 29.1 °C
Ti2 = Initial temp. of water = 25.3 °C
m1 = mass of metal = 25.605 g
Tf1 = final temp. of metal = 29.1 °C
Ti1 = initial temp. of metal = 100.5 °C

Cs1 = ( -1 * 25 g * 4.18 J/g°C * (29.1 °C - 25.3 °C) ) / ( 25.605 g * (29.1 °C - 100.5 °C) ) = 0.217 J/g°C