Answer :
Let the final temperature be "T".
For the piece of copper :
- mass, [tex]\sf{m_c=40\ g.}[/tex]
- specific heat capacity, [tex]\sf{c_c=0.4\ J\,g^{-1}\,K^{-1}.}[/tex]
- initial temperature, [tex]\sf{T_c=200^{\circ}C.}[/tex]
Then the heat of copper :
[tex]\sf{\dashrightarrow Q_c=m_cc_c\,\Delta\!T_c}[/tex]
[tex]\sf{\dashrightarrow Q_c =16(T-200)\ J}[/tex]
For copper calorimeter :
- mass, [tex]\sf{m_{cc} =60\ g.}[/tex]
- specific heat capacity, [tex]\sf{c_{cc} =0.4\ J\,g^{-1}\,K^{-1}.}[/tex]
- initial temperature, [tex]\sf{T_{cc} =25^{\circ}C.}[/tex]
Then the heat of copper calorimeter :
[tex]\sf{\dashrightarrow Q_{cc} =m_{cc}c_{cc}\,\Delta\!T_{cc}}[/tex]
[tex]\sf{\dashrightarrow Q_{cc} =24(T-25)\ J}[/tex]
For water :
- mass, [tex]\sf{m_w=50\ g. }[/tex]
- specific heat capacity, [tex]\sf{c_w= 4.2\ J\,g^{-1}\,K^{-1}.}[/tex]
- initial temperature, [tex]\sf{T_w=25^{\circ}C.}[/tex]
Then heat of water :
[tex]\sf{\dashrightarrow Q_w=m_wc_w\,\Delta\!T_w}[/tex]
[tex]\sf{\dashrightarrow Q_w=210(T-25)\ J}[/tex]
By energy conservation, the sum of all these energies should be zero as there were no heat energy change before the process, i.e.,
[tex]\sf{\dashrightarrow Q_c+Q_{cc}+Q_w=0}[/tex]
[tex]\sf{\dashrightarrow16(T-200)+24(T-25)+210(T-25)=0}[/tex]
[tex]\sf{\dashrightarrow 250T- 9050=0}[/tex]
[tex]\sf{\dashrightarrow T=36.2^{\circ}C}[/tex]
[tex] \large \underline{\underline{\boxed{\sf T=36.2^{\circ}C}}}[/tex]
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[Note: in case of considering temperature difference it's not required to convert the temperatures from [tex]\sf{^{\circ}C}[/tex] to K or K to [tex]\sf{^{\circ}C}[/tex].]
