Question

$250g$ of water at $30{}^{\circ }C$is present in a copper vessel of mass $50g$. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to $5{}^{\circ }C$.

(Specific latent heat of fusion of ice $=336\times {10}^{3}Jk{g}^{-1}$

Specific heat capacity of copper vessel $=400Jk{g}^{-1}{}^{\circ }{C}^{-1}$,

Specific heat capacity of water $=4200Jk{g}^{-1}{}^{\circ }{C}^{-1}$)

Answer 1

Step 1: Given data

mass of water ($m_w$) = $250g=0.25kg$

mass of vessel ($m_v$) $=50g=0.05kg$

change in temperature of water and vessel ($\Delta T_1$)$=25{}^{\circ }C$

change in temperature of ice ($\Delta T_2$)$=5{}^{\circ }C$

Specific latent heat of fusion of ice ($L$) $=336\times {10}^{3}Jk{g}^{-1}$
Specific heat capacity of the copper vessel ($s_v$)$=400Jk{g}^{-1}{}^{\circ }{C}^{-1}$,
Specific heat capacity of water ($s_w$)$=4200Jk{g}^{-1}{}^{\circ }{C}^{-1}$

Step 2: Find the mass of ice

mass of ice required ($m_i$)

Analyzing the amount of heat absorbed or released:

When ice is used to cool down water from $30{}^{\circ }C$to $5{}^{\circ }C$, it attains an equilibrium temperature with water (i.e., $5{}^{\circ }C$).

Thus, the total heat absorbed by ice to reach $5{}^{\circ }C$will be

$Q_2=m_{i}L+m_i{s}_w\Delta T_2$

(Heat required phase change and further increasing the temperature to $10{}^{\circ }C$)

The heat released by water and vessel to cool down from $30{}^{\circ }C$ to $5{}^{\circ }C$ will be:

$Q_1=m_v{s}_v\Delta T_1+m_w{s}_w\Delta T_1$

Using the principle of calorimeter:

According to the principle of calorimeter:

Heat absorbed = Heat released

$\Rightarrow Q_1=Q_2$

$\therefore m_v{s}_v\Delta T_1+m_w{s}_w\Delta T_1=m_{i}L+m_i{s}_w\Delta T_2$

Simplifying the above equation for the mass of ice:

$\therefore m_i=\frac{m_v{s}_v\Delta T_1+m_w{s}_w\Delta T_1}{L+s_w\Delta T_2}$

Step 3: Calculating the mass of ice

Substituting the given values in the above-obtained expression:

$\therefore m_i=\frac{0.05\times 400\times 25+0.25\times 4200\times 25}{336000+4200\times 5}$

$$\begin{gathered} \therefore m_i=\frac{500+26250}{357000} \\ \therefore m_i=0.075kg \end{gathered}$$

$\therefore m_i=75g$

Hence, $75g$ of ice is required to bring down the temperature of the vessel and its contents to $5{}^{\circ }C$