A copper calorimeter can with mass 0.446 kgcontains 0.0950 kg of ice. The system is initially at$\mathbf{0}\mathbf{.}\mathbf{0}\mathbf{°}\mathit{C}$. (a) If$\mathbf{0}\mathbf{.}\mathbf{0350}\mathit{k}\mathit{g}$of steam at$\mathit{100}\mathit{.}\mathit{0}\mathit{°}\mathit{C}$and 1.00 atmpressure is added to the can, what is the final temperature of the calorimeter can and its contents? (b) At the final temperature, how many kilograms are there of ice, how many of liquid water, and how many of steam?

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,$∆T$andis the change in temperature.

The heat transfer in phase change is,

$Q=m{L}_f$… (2)

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

(a)

The mass of the can is$m_c=0.446kg$, the mass of the ice is$m_{ice}=0.0950kg$, and the mass of the stream is$m_{stream}=0.0350kg$. The initial temperature of the ice is$T_{ice}=0.0C°$, and the initial temperature of the steam is$T_{stream}=100C°$.

Let the final temperature be T .

By substituting the value of $m_c=0.446kg$and change in temperature into formula heat energy (1), the heat energy of the can is,

$$\begin{gathered} Q=m\times c\times \delta T \\ =0.446kg\times 390J/kg.K\times \left[T-0C°\right] \\ =173.9J/k\times T \end{gathered}$$

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

Adding the equation (1) and (2), the heat energy of the ice is,

$$\begin{gathered} Q=m\times c\times \delta T+m\times L_f \\ =0.0950kg\times 4190J/kg.K\times \left[T-0C°\right]+0.0950kg\times 334\times {10}^{3}J/kg \\ =3.173\times {10}^{4}J+398J/k\times T \end{gathered}$$

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

Adding the equation (1) and (2), the heat energy of the stream is,

$$\begin{gathered} Q=m\times c\times \delta T-m\times L_f \\ =0.0350kg\times 4190J/kg.K\times \left[T-100C°\right]+0.0350kg\times 2256\times {10}^{3}J/kg \\ =-9.362\times {10}^{4}J+146.6J/k\times T \end{gathered}$$

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

$$\begin{gathered} \left[Q_{can}\right]+\left[Q_{ice}\right]+\left[Q_{steam}\right]=0 \\ {\left[173.9J/k\times T\right]}_{can}+{\left[3.173\times {10}^{4}J+398J/k\times T\right]}_{ice}+{\left[-9.362\times {10}^{4}J+146.6J/k\times T\right]}_{steam}=0 \\ T=\frac{6.189\times {10}^{4}J}{718.5J/k} \\ =86.1C° \end{gathered}$$

Hence, the final temperature is$86.1C°$.

(b)

At the final temperature, there is no ice and stream. There is only water with a mass of 0.13 kg .