Chapter 12.1, Problem 4PP

To determine

The specific heat of aluminum block .

Answer to Problem 4PP

The specific heat of aluminum block is $903.7\text{ J/kg}\text{.K}$

Explanation of Solution

Given:

Weight of container is $1.00\times {10}^2 \text{g}$ . It has a temperature of $100°\text{C}$ and is placed in $1.00\times {10}^2 \text{g}$ of water at a temperature of $10.0°\text{C}$ . The final or equilibrium temperature of the mixture is $26°\text{C}$ .

Formula Used:

When a hot object at a temperature $T_A$ (in Kelvin) is immersed in cooler water at initial temperature $T_B$ (Kelvin), the temperature of water increases and reaches an equilibrium value. This is due to the exchange of heat energy between the object and water. If the mass of the object and water is $m_A$ and $m_B$ respectively, with specific heat $C_A$ and $C_B$ , the equilibrium temperature is given by the expression,

  $T_f=\frac{m_A{C}_A{T}_A+m_B{C}_B{T}_B}{m_A{C}_A+m_B{C}_B}$ ....... (1)

By knowing the equilibrium temperature of the mixture and the remaining parameters, the above equation can be solved to obtain the specific heat of the hot object $C_A$ .

Calculation:

Here, the hot object is aluminum which has a mass of $m_A\text{=1}.00\times \text{1}{0}^{\text{2}}–\text{g}$ . The water in the container has a mass of $m_B\text{=1}.00\times \text{1}{0}^{\text{2}}–\text{g}$ and specific heat $C_B=4180\text{ J/kg}\cdot \text{K}$ .

The temperature of Aluminum in Kelvin can be calculated as,

  $\begin{array}{c}{T}_A=100+273\\ =373\text{ K}\end{array}$

Similarly, the temperature of water in Kelvin is,

  $\begin{array}{c}{T}_A=10+273\\ =283\text{ K}\end{array}$

The equilibrium temperature in Kelvin is,

  $\begin{array}{c}{T}_f=26+273\\ =299\text{ K}\end{array}$

Substituting the given parameters along with equilibrium temperature in (1),

  $\begin{array}{c}299=\frac{1.0\times {10}^2\times C_A\times 373+1.0\times {10}^2\times 4180\times 283}{1.0\times {10}^2\times C_A+1.0\times {10}^2\times 4180}\\ 299=\frac{37300C_A+118294000}{100C_A+418000}\end{array}$

Solving the equation,

  $\begin{array}{l}29900C_A+124982000=37300C_A+118294000\\ 7400C_A=6688000\\ C_A=903.7\text{ J/kg}\cdot \text{K}\end{array}$

Conclusion:

Thus, the specific heat of aluminum is $903.7\text{ J/kg}\cdot \text{K}$ .