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SSSSS [86.1K]
3 years ago
11

Hi.

Physics
1 answer:
docker41 [41]3 years ago
7 0
<h3>Answer :</h3>

Let the final temperature be "T".

For the piece of copper :

  • mass, \sf{m_c=40\ g.}

  • specific heat capacity, \sf{c_c=0.4\ J\,g^{-1}\,K^{-1}.}

  • initial temperature, \sf{T_c=200^{\circ}C.}

Then the heat of copper :

\sf{\dashrightarrow Q_c=m_cc_c\,\Delta\!T_c}

\sf{\dashrightarrow Q_c =16(T-200)\ J}

For copper calorimeter :

  • mass, \sf{m_{cc} =60\ g.}

  • specific heat capacity, \sf{c_{cc} =0.4\ J\,g^{-1}\,K^{-1}.}

  • initial temperature, \sf{T_{cc} =25^{\circ}C.}

Then the heat of copper calorimeter :

\sf{\dashrightarrow Q_{cc} =m_{cc}c_{cc}\,\Delta\!T_{cc}}

\sf{\dashrightarrow Q_{cc} =24(T-25)\ J}

For water :

  • mass, \sf{m_w=50\ g. }

  • specific heat capacity, \sf{c_w= 4.2\ J\,g^{-1}\,K^{-1}.}

  • initial temperature, \sf{T_w=25^{\circ}C.}

Then heat of water :

\sf{\dashrightarrow Q_w=m_wc_w\,\Delta\!T_w}

\sf{\dashrightarrow Q_w=210(T-25)\ J}

By energy conservation, the sum of all these energies should be zero as there were no heat energy change before the process, i.e.,

\sf{\dashrightarrow Q_c+Q_{cc}+Q_w=0}

\sf{\dashrightarrow16(T-200)+24(T-25)+210(T-25)=0}

\sf{\dashrightarrow 250T- 9050=0}

\sf{\dashrightarrow T=36.2^{\circ}C}

\large \underline{\underline{\boxed{\sf T=36.2^{\circ}C}}}

<u>____________________________</u>

[Note: in case of considering temperature difference it's not required to convert the temperatures from \sf{^{\circ}C} to K or K to \sf{^{\circ}C}.]

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