A Styrofoam bucket of negligible mass contains 1.75kgof water and 0.450kgof ice. More ice, from a refrigerator at$\mathbf{-}\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{°}\mathit{C}$, is added to the mixture in the bucket, and when thermal equilibrium has been reached, the total mass of ice in the bucket is 0.884 kg. Assuming no heat exchange with the surroundings, what

mass of ice was added?

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

$Q=mc∆T$

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,

$Q=m{L}_f$

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

The mass of the water is $m_w=1.75kg$, the mass of the ice is$m_{ice}=0.45kg$, and the total mass of the ice in the bucket is$m_t=0.884kg$. The initial temperature of the ice is$T_{ice}=-15C°$.

Note that to increase the mass of the ice, the part of water is also converted into ice.

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

$$\begin{gathered} {\left[m\times C\times ∆T\right]}_{ice}+{\left[-m\times L_f\right]}_w=0 \\ {\left[m\times C\times ∆T\right]}_{ice}={\left[m\times L_f\right]}_w.......\left(1\right) \end{gathered}$$

Here, ${\left[m\times C\times ∆T\right]}_{ice}$is the heat energy of the ice due to change in temperature, and ${\left[-m\times L_f\right]}_w$is the heat energy due to phase transition in the conversion of water into ice.

Since the mass of the ice in the container increases by 0.434kg , then the mass of the added ice and the mass of the converted water is equal to 0.434kg .

$$\begin{gathered} m_{ice}+m_w=0.434 \\ {m}_w=0.434-m_{ice} \end{gathered}$$

Substitute the value of$m_w=0.434-m_{ice}$in the equation (1) to get the value of.

$$\begin{gathered} {\left[m\times c\times \delta T\right]}_{ice}={\left[\left[0.434-m_{ice}\right]\times L_f\right]}_w \\ {m}_{ice}=\frac{0.434\times L_f}{c\times \delta T+L_f} \\ =\frac{0.434\times 334-{10}^{3}J/kg}{2100J/kg.K\times \left(15\right)C°+334\times {10}^{3}J/kg} \\ =0.396kg \end{gathered}$$

Hence, the mass of the added ice is 0.396kg .