Can someone help me with this problem.... i have an exam coming tomorrow and im doing practice problems but i dont know how to do this

Calculate the specific heat of a metal if 2.36102 grams of it at 99.5 degrees celsius is added to 125.0 mL of water at 22.0 degrees celsius. The final temperature of the system is 25.4 degrees celsius.

Use Σq = 0 = (m∙c∙ΔT)water + (m∙c∙ΔT)metal

Water
m = 125ml = 125g <=> (1ml H₂O = 1gm H₂O)
c = 1cal/g∙⁰C or 4.184j/g∙⁰C
ΔT = 25.4⁰C – 22.0⁰C
Metal
m = 2.36102g
c = ?
ΔT = 99.5⁰C – 22.0⁰C
(m∙c∙ΔT)water + (m∙c∙ΔT)metal = 0
[(125)(1)(25.4 – 22.0)] + [(2.36102)(c)(25.4 – 99.5)] = 0
Solve for ‘c’.

Oops! ΔT(metal) = 25.4⁰C – 99.5⁰C

To calculate the specific heat of a metal, you can use the equation:

q = m * c * ΔT

Where:
- q is the heat transferred
- m is the mass of the metal
- c is the specific heat of the metal
- ΔT is the change in temperature

First, let's calculate the heat transferred to the water. We'll assume that the specific heat capacity of water is 4.18 J/g°C.

q_water = m_water * c_water * ΔT_water

m_water = 125.0 mL (the density of water is 1 g/mL, so the mass is equal to volume)
ΔT_water = final temperature - initial temperature = 25.4°C - 22.0°C

Now, calculate the heat transferred to the metal. We'll assume the mass of the metal is given as 2.36102 grams.

q_metal = m_metal * c_metal * ΔT_metal

m_metal = 2.36102 grams
ΔT_metal = final temperature - initial temperature = 25.4°C - 99.5°C

The total heat transferred to the system is zero since the system is isolated.

q_total = 0

Now, set up the equations and solve for c_metal:

m_water * c_water * ΔT_water + m_metal * c_metal * ΔT_metal = q_total

Insert the values and solve for c_metal:

125.0 mL * 1 g/mL * 4.18 J/g°C * (25.4°C - 22.0°C) + 2.36102 g * c_metal * (25.4°C - 99.5°C) = 0

Simplify and solve for c_metal.