Question

A block of copper (one mole has a mass of 63.5 g ) at a temperature of $\mathbf{35}\mathbf{°}$is put in contact with a 100 gblock of aluminum (molar mass 27 g) at a temperature of$\mathbf{20}\mathbf{°}\mathbf{}\mathbf{C}$. The blocks are inside an insulated enclosure, with little contact with the walls. At these temperatures, the high-temperature limit is valid for the specific heat. Calculate the final temperature of the two blocks. Do NOT look up the specific heats of aluminum and copper; you should be able to figure them out on your own.

Short answer

The final temperature of two blocks is $37°\mathrm{C}$.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The mass of copper block is, $m_{cu}=50\mathrm{g}$.
• The temperature of the block of copper is, $T_{cu}=35°\mathrm{C}=308\mathrm{K}$.
• The mass of aluminium block is, $m_{AI}=100\mathrm{g}$.
• The temperature of the block of aluminium is, $T_{AI}=20°\mathrm{C}=293\mathrm{K}$.

Concept/Significance of specific heat capacity

The amount of heat absorbed by a unit mass of an item to produce a unit temperature rise is referred to as its specific heat capacity.

Determination of the final temperature of the two blocks

The specific heat of a metal is given by,

$C=\left(\frac{m}{M}{N}_A\right)\left(3k_B\right)$

Here, m is the given mass of the atom and Mis the molar mass of the atom,$N_A$is the Avogadro number whose value is $6.023\times {10}^{23}\mathrm{J}/\mathrm{K}$.

The heat lost by copper block is equal to Aluminum is given by,

$$\begin{gathered} Q_{cu}=Q_{AI} \\ {m}_{cu}{C}_{cu}\left(T_{cu}-T\right)=m_{AI}{C}_{AI}\left(T_{AI}-T\right) \end{gathered}$$

The total temperature is given by,

$$\begin{gathered} T=\frac{m_{AI}{C}_{AI}{T}_{AI}-m_{cu}{C}_{cu}{T}_{cu}}{m_{AI}{C}_{AI}-m_{cu}{C}_{cu}} \\ =\frac{\left({\displaystyle \frac{{m_{AI}}^2}{m_{AI}}}\right)T_{AI}+\left({\displaystyle \frac{{m_{cu}}^2}{m_{cu}}}\right)T_{cu}}{{\displaystyle \frac{{M_{AI}}^2}{M_{AI}}}+{\displaystyle \frac{{m_{cu}}^2}{m_{cu}}}} \end{gathered}$$

Substitute all the values in the above,

localid="1657860339433" $$\begin{gathered} T=\frac{T_{cu}/63.5+4T_{AI}/27}{\left(4/27+1/63.5\right)} \\ =\frac{308K/63.5+293\mathrm{K}}{\left(4/27+1/63.5\right)} \\ =309.89\mathrm{K} \end{gathered}$$

The final temperature of two blocks is given by,

$$\begin{gathered} T_F=\left(309.89-273\right)°\mathrm{C} \\ =37°\mathrm{C} \end{gathered}$$

Thus, the final temperature of two blocks is $37°\mathrm{C}$.