Question

A $2.50\text{kg}$lump of aluminum is heated to$\mathbf{92}\mathbf{.0}\mathbf{°}\mathbf{\text{C}}$and then dropped into$\mathbf{8}\mathbf{.00}\mathbf{}\mathbf{\text{kg}}$of water at$\mathbf{5}\mathbf{.00}\mathbf{°}\mathbf{\text{C}}$. Assuming that the lump–water system is thermally isolated, what is the system’s equilibrium temperature?

Short answer

The equilibrium temperature of the system is${10.5}^0\text{C}.$

Step-by-step solution

Identification of given data

i) Mass of aluminum$m_A=2.50\text{kg}$

ii) Initial temperature of aluminum$T_{iA}={92}^0\text{C}$

iii) Initial temperature of water$T_{iw}={5.00}^0\text{C}$

iv) Mass of water,$m_w=8.00\text{kg}$

v) Thermal conductivity of aluminum,$k_{Al}=900{\text{J/kg.}}^0\text{C}$

vi) Thermal conductivity of water,$k_w=4186.8{\text{J/kg.}}^0\text{C}$.

Understanding the concept of heat transfer

The heat transfer from the hotter end to the cooler end due to the contact of two substances is called heat transfer by conduction. Thus, here, when aluminum is brought in contact with the surface of the water, it transfers some heat which is equal to the heat transferred by the water surface to the aluminum surface within the isolated system. As the system is thermally isolated, its total energy remains the same. Therefore, we can write equations for aluminum and water using the molar heat capacity. Then, by equating these equations, we can find the equilibrium temperature.

Formula:

Heat transferred by the body via thermal radiation, $Q=mc\Delta T$ …(i)

where,$m$ is the mass of the substance, $c$is the specific heat of the substance, $\Delta T$and is the temperature difference.

Determining the equilibrium temperature

Ifis the final equilibrium temperature of the aluminum water system, then the energy transferred from the aluminum using equation (i) $Q_A=m_A{c}_A(T_{iA}-T_f)$is.

Similarly, the energy transferred into water as heat using equation (i) is

$Q_w=m_w{c}_w(T_f-T_{iw})$

Therefore, at equilibrium

$\begin{array}{c}{Q}_A=Q_w\\ m_A{c}_A(T_{iA}-T_f)=m_w{c}_w(T_f-T_{iw})\\ m_A{c}_A{T}_{iA}-m_A{c}_A{T}_f=m_w{c}_w{T}_f-m_w{c}_w{T}_{iw}\\ m_A{c}_A{T}_{iA}+m_w{c}_w{T}_{iw}=m_w{c}_w{T}_f+c_A{T}_f\\ T_f=\frac{m_A{c}_A{T}_{iA}+m_w{c}_w{T}_{iw}}{m_w{c}_w+m_A{c}_A}\end{array}$

Substituting the given values in the above equation, we get the equilibrium temperature as

$\begin{array}{c}{T}_f=\frac{(2.50\text{kg})\left(900{\text{J/kg.}}^{\text{0}}\text{C}\right)\left({92}^{\text{0}}\text{C}\right)+(8.00\text{kg})(4186.8{\text{J/kg.}}^{\text{0}}\text{C})\left({5.0}^0\text{C}\right)}{(8.00\text{kg})(4186.8{\text{J/kg.}}^{\text{0}}\text{C})+(2.50\text{kg})(900{\text{J/kg.}}^{\text{0}}\text{C})}\\ ={10.5}^0\text{C}\end{array}$

Therefore, the equilibrium temperature of the system is${10.5}^0\text{C}.$