Solved Problem on Heat
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A block of mass m1, specific heat c1, and temperature T1 is brought into contact with a block of another material, of mass, specific heat, and temperature, respectively, m2, c2, and T2. After reaching thermal equilibrium between the blocks, if c1 and c2 are constants, and assuming that the heat transfer with the environment is negligible, calculate the final equilibrium temperature T.


Problem data:
  • m1, c1 and T1, mass, specific heat, and initial temperature of block 1;
  • m2, c2 and T2, mass, specific heat, and initial temperature of block 2.
Solution

We want to find the equilibrium temperature teq = T. When the blocks are brought into contact, the colder block gains heat from the hotter one and increases the temperature, the hottest block loses heat to the colder one, and the temperature decreases until they both reach the same temperature (equilibrium temperature), the heat transfer equation is given by
\[ \bbox[#99CCFF,10px] {Q=mc\left(t_{eq}-t_{0}\right)} \]
Writing this expression for each of the blocks
\[ \begin{gather} Q_{1}=m_{1}c_{1}\left(T-T_{1}\right)\\[10pt] Q_{2}=m_{2}c_{2}\left(T-T_{2}\right) \end{gather} \]
The problem tells us that the heat transfer between the system and the environment is negligible, so the system is isolated and there is only heat transfer between the blocks, as heat is energy transferred, we can use the conservation of energy, "the sum of the heat transferred is zero in a thermally insulated system".
\[ \begin{gather} \sum Q=0\\[5pt] Q_{1}+Q_{2}=0\\[5pt] m_{1}c_{1}\left(T-T_{1}\right)+m_{2}c_{2}\left(T-T_{2}\right)=0\\[5pt] m_{1}c_{1}T-m_{1}c_{1}T_{1}+m_{2}c_{2}T-m_{2}c_{2}T_{2}=0\\[5pt] m_{1}c_{1}T+m_{2}c_{2}T=m_{1}c_{1}T_{1}+m_{2}c_{2}T_{2}\\[5pt] T\left(m_{1}c_{1}+m_{2}c_{2}\right)=m_{1}c_{1}T_{1}+m_{2}c_{2}T_{2} \end{gather} \]
\[ \bbox[#FFCCCC,10px] {T=\frac{m_{1}c_{1}T_{1}+m_{2}c_{2}T_{2}}{m_{1}c_{1}+m_{2}c_{2}}} \]
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