Answer:
The final temperature is 61.65 °C
Explanation:
mass of copper pot [tex]m_{c}[/tex] = 2 kg
temperature of copper pot [tex]T_{c}[/tex] = 20 °C  (the pot will be in thermal equilibrium with the room)
specific heat capacity of copper [tex]C_{c}[/tex]= 385 J/kg-°C
The heat content of the copper pot = [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{c}[/tex] = 2 x 385 x 20 = 15400 J
mass of boiling water [tex]m_{w}[/tex] = 200 g = 0.2 kg
temperature of boiling water [tex]T_{w}[/tex] = 100 °C
specific heat capacity of water [tex]C_{w}[/tex] = 4182 J/kg-°C
The heat content of the water = [tex]m_{w}[/tex][tex]C_{w}[/tex][tex]T_{w}[/tex] = 0.2 x 4182 x 100 = 83640 J
The total heat content of the water and copper mix [tex]H_{T}[/tex] = 15400 + 83640 = 99040 J
This same heat is evenly distributed between the water and copper mass to achieve thermal equilibrium, therefore we use the equation
[tex]H_{T}[/tex] = Â [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{f}[/tex] + [tex]m_{w}[/tex][tex]C_{w}[/tex]
where [tex]T_{f}[/tex] is the final temperature of the water and the copper
substituting values, we have
99040 = (2 x 385 x [tex]T_{f}[/tex]) + (0.2 x 4182 x
99040 = 770[tex]T_{f}[/tex] + 836.4
99040 = 1606.4[tex]T_{f}[/tex]
[tex]T_{f}[/tex] = 99040/1606.4 = 61.65 °C
