Question

80gm of water at 30^oC is poured on a large block of ice at0^oC .The mass of ice that melts is ?

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Heat}\mathrm{lost}\mathrm{by}\mathrm{water}\mathrm{during}\mathrm{cooling}\mathrm{from}80°\mathrm{C}\mathrm{to}0°\mathrm{C}, \\ ∆\mathrm{H}=\mathrm{ms}∆\mathrm{T} \\ ∆\mathrm{H}=30\times 1\times \left(80-0\right) \\ ∆\mathrm{H}=2400\mathrm{cal} \\ \mathrm{Latent}\mathrm{heat}\mathrm{of}\mathrm{fusion}\mathrm{of}\mathrm{ice},\mathrm{L}=80\mathrm{cal}/\mathrm{gm} \\ \mathrm{mass}\mathrm{of}\mathrm{ice}\mathrm{that}\mathrm{melts},\mathrm{m}\text{'}=\frac{∆\mathrm{H}}{\mathrm{L}} \\ \mathrm{m}\text{'}=\frac{2400}{80}=30\mathrm{gm} \end{gathered}$$