19 g of water at 30⁰C and 5 g of ice at $-$20⁰C are mixed together in a calorimeter. What is the final temperature of the mixture? (Take, specific heat of ice = 0.5 cal ${\mathrm{g}}^{-1}$ ⁰$\mathrm{C}{ }^{-1}$ and latent heat of fusion of ice = 80 cal ${\mathrm{g}}^{-1}$)

detailed solution

Correct option is C

Here, specific heat of ice,  cice =0.5calg−1oC−1Specific heat of water, cwater =1calg−1oC−1Latent heat of fusion of ice ,Lice =80calg−1Here, ice will absorb heat while hot water will release it.Let T be the final temperature of the mixture.Assuming water equivalent of calorimeter to be neglected.Heat given by water,   Q1=mwater cwater ΔT=19×1×(30−T)=570−19T …(i)Heat absorbed by ice,Q2=mice ×cice ×[0−(−20)]+mice ×Lf ice  +mice ×cwater ×(T−0)=5×0.5×20+5×80+5×1×T=450+5T… (ii) According to principle of calorimetry,  Q1=Q2570−19T=450+5T⇒T=12024=5∘C