Question
100 g of ice at −10°C is mixed with 300 g of water at 50°C in an insulated calorimeter (calorimeter heat capacity negligible). Find the final temperature and final composition. Take cice=0.5 cal/g °C, cwater=1 cal/g °C, Lf=80 cal/g.
Solution — Step by Step
Heat to warm ice from −10°C to 0°C: Q1=miciΔT=100×0.5×10=500 cal.
Heat to melt all of the ice at 0°C: Q2=miLf=100×80=8000 cal.
Total to fully melt: Q1+Q2=8500 cal.
Maximum heat the water can release as it cools from 50°C to 0°C:
Qw=mwcwΔT=300×1×50=15,000 cal.
Since Qw>Q1+Q2, all the ice will melt, and we have leftover heat to warm the resulting water. Final T>0°C.
Let final temperature = T. Heat lost by hot water = heat gained by ice (warming + melting + heating after melting).
300×1×(50−T)=500+8000+100×1×(T−0)
15000−300T=8500+100T
6500=400T⟹T=16.25°C.
Final answer: Final T=16.25°C, all 400 g becomes liquid water.
Why This Works
The strategy in mixing problems: first check whether all the ice melts. Compare the heat needed to fully melt ice vs the maximum heat available from cooling the warmer substance to 0°C. Three cases:
- Ice doesn’t fully melt → final T=0°C, with some ice + water.
- Ice fully melts and final T>0°C — our case.
- Water freezes onto the ice → final T<0°C (rare).
Skipping this check is the classic trap.
Alternative Method
Compute net heat budget: Qw−Q1−Q2=15000−500−8000=6500 cal available to warm the merged 400 g of water from 0°C. So T=6500/(400×1)=16.25°C. Same answer in fewer lines.
Common Mistake
Students often forget Q1 — the heat to warm ice from below 0°C to 0°C before melting begins. Just plugging in Lf×mi misses 500 cal. Always trace the full path: subzero ice → 0°C ice → 0°C water → final T.