$$19$$ 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$$ cal $$g-1$$ $$0C$$ $$-1$$ and latent heat of fusion of ice $$=$$ $$80$$ cal $$g-1)

Question

$$19$$ 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$$ cal $$g-1$$ $$0C$$ $$-1$$ and latent heat of fusion of ice $$=$$ $$80$$ cal $$g-1)

Infinity Learn · Accepted Answer

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 Δ$$T=19$$×$$1$$×(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

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 g-1 0C -1 and latent heat of fusion of ice = 80 cal g-1)

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