91Ó°ÊÓ

The given data is listed below:

• The mass of the aluminium block is,$m_{Al}=1.5 \text{kg}$
• The mass of the copper block is,$m_{Cu}=2.1 \text{kg}$
• The initial temperature for aluminium block is,$T_{i-Al}=45°\text{C}$
• The initial temperature for copper block is,$T_{i-Cu}=18°\text{C}$
• The specific heat of aluminium is,$C_{Al}=0.91 \text{J/g}$
• The specific heat of copper is,$C_{Cu}=0.39 \text{J/g}$

The energy transferred can be determined by taking the product of mass of the object, temperature of the object, and the specific heat of the object. It can be expressed as follows:

$\mathbf{Q}\mathbf{=}\mathbf{³¾°äΔ°Õ}$…(1)

Here, m is the mass of the object, C is the specific heat of the object, and $\mathrm{Δ}T=T_f-T_i$is the temperature difference of the object ($T_i$is the initial temperature of the object and is the final temperature of the object).

It is known that the total energy transfer is zero. Write the expression for the energy transfer by two blocks and use equation (1).

$$\begin{gathered} \mathrm{Δ}{Q}_{Al}+\mathrm{Δ}{Q}_{Cu}=0 \\ \left(m_{Al}{C}_{Al}\mathrm{Δ}{T}_{Al}\right)+\left(m_{Cu}{C}_{Cu}\mathrm{Δ}{T}_{Cu}\right)=0 \\ \left(m_{Al}{C}_{Al}\left(T_f-T_{i-Al}\right)\right)+\left(m_{Cu}{C}_{Cu}\left(T_f-T_{i-Cu}\right)\right)=0 \end{gathered}$$

Here,$T_f$ is the final temperature of both the blocks.

Rearrange the above expression and substitute all the values in the above expression.

$$\begin{gathered} T_f=\frac{\left(m_{Al}{C}_{Al}{T}_{i-Al}\right)+\left(m_{Cu}{C}_{Cu}{T}_{i-Cu}\right)}{\left(m_{Al}{C}_{Al}\right)+\left(m_{Cu}{C}_{Cu}\right)} \\ =\frac{\left(\left(1.5 \text{kg}\right)\left(0.91 \text{J/g}\right)\left(45°\text{C}\right)\right)+\left(\left(2.1 \text{kg}\right)\left(0.39 \text{J/g}\right)\left(18°\text{C}\right)\right)}{\left(\left(1.5 \text{kg}\right)\left(0.91 \text{J/g}\right)\right)+\left(\left(2.1 \text{kg}\right)\left(0.39 \text{J/g}\right)\right)} \\ =34.9°\text{C} \end{gathered}$$

Thus, the final temperature of both the blocks is $34.9°\text{C}$.