Chemistry homeowrk help

Question

Calculate the specific heat of copper. The specific heat of water is 4.18 J/(g·°C)

the mass of water and the mass of copper.

M water=60

M copper= 17.920

The temperature change of the water and the copper.

Δ𝑇water = 0.78

Δ𝑇copper = -28.21

Answer 1

Looks like the experiment was to heat the copper and drop it into water and record the temperature change. Assuming that is the case, we can approach the problem as follows:

heat lost by copper = heat gained by water

heat lost by copper = q = mC∆T

q = (17.920 g)(C_Cu)(28.21)

heat gained by water = q = mC∆T

q = (60 g)(4.184 J/g/deg)(0.78 deg) = 195.8 J Since this is the heat gained by water it must be the heat lost by the copper. Thus...

195.8 J = (17.920 g)(C_Cu)(28.21 deg)

C_Cu = 195.8 J/(17.920)(28.21)

C_Cu = 0.387 J/g/º = specific heat of copper