91Ó°ÊÓ

The heat required to change temperature of a certain mass is given by,

$Q=mc∆T$… (1)

Here, m is the mass of the material, c is the specific heat of the material, and$∆T$is the change in temperature.

The heat transfer in phase change is,

$Q=m{L}_l$… (2)

Here, is the mass of the material, and $L_f$is the latent heat for the phase change.

The mass of the water is $m_w=2kg$, the mass of the aluminum mug is $m_{mug}=0.257kg$, and the mass of the ice is $m_{ice}=0.120kg$. The initial temperature of the ice is $T_{ice}=-15°C$, and the initial temperature of the mug and the soft drink is$T_w=T_{mug}=20°C$.

Let the final temperature be T .

By substituting the value of $m_w=2kg$and change in temperature into formula heat energy (1), the heat energy of the water is,

$$\begin{gathered} Q=m×c×\mathrm{δ}T \\ =2kg×4190J/kg.K×\left[T-20C°\right] \\ =8380.\left[T-20°C\right] \end{gathered}$$

By substituting the value of $m_{mug}=0.257kg$and change in temperature into formula heat energy (1), the heat energy of the mug is,

$$\begin{gathered} Q=m×c×\mathrm{δ}T \\ =0.257kg×910J/kg×\left[T-20°C\right] \\ =233.87.\left[T-20°C\right] \end{gathered}$$

There are two heat energy for ice. One is due to temperature change and the another is due to phase transition.

By substituting the value of$m_{ice}=0.120kg$and change in temperature into formula heat energy (1), the heat energy of the ice due to temperature change is,

$$\begin{gathered} Q=m×c×\mathrm{δ}T \\ =0.120kg×2100J/kg.K×\left[T-\left(-15\right)C°\right] \\ =252.\left[T+15C°\right] \end{gathered}$$

By substituting the value of$m_{ice}=0.120kg$and change in temperature into formula heat energy (2), the heat energy of the ice due to phase transition is,

$$\begin{gathered} Q=m×L_f \\ =0.120kg×334×{10}^{3}J/kg \\ =40080J \end{gathered}$$

The total heat energy of the system at thermal equilibrium is zero.

$$\begin{gathered} \left[T-20C°\right].\left[8380+233.87\right]+252.\left[T+15C°\right]+40080J=0 \\ T=14.48C° \end{gathered}$$

Hence, the final temperature is$14.48C°$.