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".
\[
\begin{gather}
\sum Q=0\\[5pt]
Q_{1}+Q_{2}=0\\[5pt]
m_{1}c_{1}\left(T-T_{1}\right)+m_{2}c_{2}\left(T-T_{2}\right)=0\\[5pt]
m_{1}c_{1}T-m_{1}c_{1}T_{1}+m_{2}c_{2}T-m_{2}c_{2}T_{2}=0\\[5pt]
m_{1}c_{1}T+m_{2}c_{2}T=m_{1}c_{1}T_{1}+m_{2}c_{2}T_{2}\\[5pt]
T\left(m_{1}c_{1}+m_{2}c_{2}\right)=m_{1}c_{1}T_{1}+m_{2}c_{2}T_{2}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{T=\frac{m_{1}c_{1}T_{1}+m_{2}c_{2}T_{2}}{m_{1}c_{1}+m_{2}c_{2}}}
\]