Since the heat gained by the water in equals the heat lost, calculate the specific heat of copper.

Question

mass of calorimeter (g)	2.50
mass of calorimeter & water(g)	84.02
mass of water (g)	81.52
mass of metal (g)	4.42
Ti of cool water (℃)	21.2
Ti of metal (℃)	101.0
Tf of water (℃)	22.0
Tf of metal (℃)	22.0
ΔT of water (℃)	0.8
ΔT of metal (℃)	-79

q(energy lost/gained)=273J

Answer 1

heat gained by water = q = mC∆T

m = 84.02 g - 2.50 g = 81.52 g

C = 4.184 J/gº

∆T = 22.0 - 21.2 = 0.8º

q = (81.52 g)(4.184 J/gº)(0.8º) = 273 J

heat gained by copper = 273 J = mC∆T

m = 4.42 g

C = ?

∆T = 101 - 22 = 79º

C = q / (m)(∆T) = 273 J / (4.42g)(79º)

C = 0.782 J/gº = specific heat of copper

Answer 2

Equation:

Q = (m) (Cp) (△T) 

Q = heat flow

m = mass

C_p = specific heat

△T = temp change

Equation rewritten to solve for C_p:

Cp = Q / (m x △T)

**I am not sure which number to use for each part of this equation, but this is the steps you should follow to get the answer here. Also, below is a document that goes into further detail on how to do this. Sorry I could not be of more help.

Document: https://www.sciencegeek.net/Chemistry/chempdfs/SpecificHeatCopper.pdf