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

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}}}`

____________________________

[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}`.]