Question

Calculate the amount of ice that is required to cool 150 g of water contained in a vessel of mass of 100 g at $30°\mathrm{C}$, such that the final temperature of the mixture is $5°\mathrm{C}$. Take specific heat capacity of material of vessel as $0.4\mathrm{J}\mathrm{g}°{\mathrm{C}}^{-1}$; specific latent heat of fusion of ice is $336\mathrm{J}{\mathrm{g}}^{-1}$; specific heat capacity of water is $4.2\mathrm{J}{\mathrm{g}}^{-1}°{\mathrm{C}}^{-1}$

Answer 1

Calculation of the amount of ice

Step 1: Calculating the heat lost by the vessel and water

$$\begin{gathered} \mathrm{Heat}\mathrm{lost}=\mathrm{m}\times \mathrm{c}\times ∆\mathrm{T} \\ \mathrm{Where},\mathrm{m}=\mathrm{mass}\mathrm{of}\mathrm{substance} \\ \mathrm{c}=\mathrm{specific}\mathrm{heat}\mathrm{capacity} \\ ∆\mathrm{T}=\mathrm{Temperature}\mathrm{difference} \end{gathered}$$

Heat lost by the vessel and water=$[150\mathrm{g}\mathrm{x}4.2\mathrm{J}{\mathrm{g}}^{-1}{}^{\mathrm{o}}\mathrm{C}^{-1}\mathrm{x}(30–5){}^{\mathrm{o}}\mathrm{C}+100\mathrm{g}\mathrm{x}0.4\mathrm{J}{\mathrm{g}}^{-1}{}^{\mathrm{o}}\mathrm{C}^{-1}\mathrm{x}(30-5\left){}^{\mathrm{o}}\mathrm{C}\right]\mathrm{J}$
Heat lost by the vessel and water = $(150\mathrm{x}4.2\mathrm{x}25+100\mathrm{x}0.4\mathrm{x}25)\mathrm{J}$
Heat lost by the vessel and water = $(15750+1000)\mathrm{J}$
Heat lost by the vessel and water = 16,750 J

Step 2: Calculating the heat gained by ice to change into water
Heat gained by ice changes into the water at $0°\mathrm{C}$ = $\mathrm{m}\mathrm{x}\mathrm{specific}\mathrm{latent}\mathrm{heat}\mathrm{of}\mathrm{fusion}\mathrm{of}\mathrm{ice}$
Heat gained by ice changes into the water at $0°\mathrm{C}$ = $\mathrm{m}\mathrm{x}336\mathrm{J}{\mathrm{g}}^{-1}$
Heat gained by water to come to final temperature ($5°\mathrm{C}$) = $\mathrm{m}\mathrm{x}\mathrm{specific}\mathrm{heat}\mathrm{capacity}\mathrm{of}\mathrm{water}\mathrm{x}\mathrm{final}\mathrm{temperature}$
Heat gained by water to come to final temperature ($5°\mathrm{C}$) = ($\mathrm{m}\mathrm{x}4.2\mathrm{J}{\mathrm{g}}^{-1}{}^{\mathrm{o}}\mathrm{C}^{-1}\mathrm{x}5{}^{\mathrm{o}}\mathrm{C}$)
Heat gained by water to come to final temperature ($5°\mathrm{C}$) = $21\mathrm{m}\mathrm{J}{\mathrm{g}}^{-1}$

Step 3: Calculation of the amount of ice required to cool 150 g of water
If there is no loss of heat to the surroundings
According to the principle of calorimetry:
Heat gained by ice to melt and come to final temperature = Heat lost by vessel and water
$$\begin{gathered} (336\mathrm{m}+21\mathrm{m})\mathrm{J}{\mathrm{g}}^{-1}=16750\mathrm{J} \\ \left(357\mathrm{m}\right){\mathrm{g}}^{-1}=16750 \\ \mathrm{m}=\frac{16750}{357}\mathrm{g} \\ \mathrm{m}=46.92\mathrm{g} \end{gathered}$$

Therefore the amount of ice that is required to cool 150 g of water is 46.92 g.