Question

Hi.
I need help with this question :

A piece of copper of mass 40g at 200°C is placed in a copper calorimeter of mass 60g containing 50g of water at 25°C. Ignoring heat losses, what will be the final temperature of the mixture.
(Specific heat capacity of copper= 0.4J/g/K, specific heat capacity of water= 4.2J/g/K )?

Please show workings.​

Answer 1

Answer :

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.,

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

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

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

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

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

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