Question

A vessel of negligible heat capacity contains 40 g of ice in it at 0 ºC. 8 g of steam at 100 ºCis passed into the ice to melt it. Find the final temperature of contents of the vessel.

[Specific latent heat of vaporisation of steam = 2268 J/g,
Specific latent heat of fusion of ice = 336 J/g,
Specific heat capacity of water = 4.2 J/g]

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Let}\mathrm{final}\mathrm{temperature}\mathrm{of}\mathrm{mixture}\mathrm{is}\mathrm{\theta }. \\ \mathrm{mass}\mathrm{of}\mathrm{steam},\mathrm{m}=8\mathrm{g} \\ \mathrm{mass}\mathrm{of}\mathrm{ice},\mathrm{m}\text{'}=40\mathrm{g} \\ \mathrm{Latent}\mathrm{heat}\mathrm{of}\mathrm{vapourisation}\mathrm{of}\mathrm{steam},\mathrm{L}=2268\mathrm{J}/\mathrm{g} \\ \mathrm{Latent}\mathrm{heat}\mathrm{of}\mathrm{fusion}\mathrm{of}\mathrm{ice},\mathrm{L}\text{'}=336\mathrm{J}/\mathrm{g} \\ \mathrm{specific}\mathrm{heat}\mathrm{capacity}\mathrm{of}\mathrm{water}=4.2\mathrm{J}/\mathrm{g} \\ \mathrm{Heat}\mathrm{given}\mathrm{by}\mathrm{system}=\mathrm{Heat}\mathrm{taken}\mathrm{by}\mathrm{the}\mathrm{ice} \\ \mathrm{mL}+\mathrm{mC}∆\mathrm{\theta }=\mathrm{m}\text{'}\mathrm{L}\text{'}+\mathrm{m}\text{'}\mathrm{C}\text{'}∆\mathrm{\theta }\text{'} \\ 8\times 2268+8\times 4.2\times \left(100-\mathrm{\theta }\right)=40\times 336+40\times 4.2\times \left(\mathrm{\theta }-0\right) \\ 18144+3360-33.6\times \mathrm{\theta }=13440+168\times \mathrm{\theta } \\ 201.6\times \mathrm{\theta }=8064 \\ \mathrm{\theta }=40°\mathrm{C} \end{gathered}$$