Calorimetry: Speed-Solving Techniques (2)

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Question

100100 g of ice at 10-10^\circC is mixed with 300300 g of water at 5050^\circC in an insulated container. Find the final temperature. Use cice=0.5c_{\text{ice}} = 0.5 cal/g·°C, cwater=1c_{\text{water}} = 1 cal/g·°C, latent heat of fusion L=80L = 80 cal/g.

Solution — Step by Step

Heat to warm ice from 10-10 to 00^\circC: Q1=100×0.5×10=500Q_1 = 100 \times 0.5 \times 10 = 500 cal.

Heat to melt ice at 00^\circC: Q2=100×80=8000Q_2 = 100 \times 80 = 8000 cal.

Total to bring ice to 00^\circC water: Q1+Q2=8500Q_1 + Q_2 = 8500 cal.

Heat released by 300300 g water cooling from 5050 to 00^\circC: Q3=300×1×50=15000Q_3 = 300 \times 1 \times 50 = 15000 cal.

Since 15000>850015000 > 8500, all the ice melts and we still have 150008500=650015000 - 8500 = 6500 cal of “leftover” heat.

The melted ice is now 100100 g of water at 00^\circC, joining the 300300 g already cooled to some temperature. Total water mass = 400400 g. Let final temperature be TT.

Heat balance from the equilibrium state: the leftover 65006500 cal warms 400400 g from 00^\circC:

6500=400×1×TT=16.25C6500 = 400 \times 1 \times T \Rightarrow T = 16.25^\circ\text{C}

Final answer: T16.25T \approx 16.25^\circC.

Why This Works

Calorimetry is bookkeeping: heat lost by hot stuff equals heat gained by cold stuff (in an insulated system). The catch is that “heat gained” includes warming, melting, vaporising, and cooling — each with its own formula.

The “check whether all the ice melts” step is non-negotiable. If the warm water cannot supply enough heat for the full melt, the final temperature stays at 00^\circC with some ice unmelted.

Alternative Method

Set up the equation in one shot. Let TT be the final temperature (assume all ice melts). Heat gained by ice = 100[0.5×10+80+1×T]100[0.5 \times 10 + 80 + 1 \times T]. Heat lost by water = 300×1×(50T)300 \times 1 \times (50 - T). Equate:

8500+100T=15000300T8500 + 100T = 15000 - 300T, giving 400T=6500400T = 6500, T=16.25T = 16.25^\circC. Same answer.

Always do the “is there enough heat?” check before assuming a uniform final temperature. NEET loves to set 5050 g ice with 5050 g water and trick you into a positive TT when the answer is actually 00^\circC with partial ice remaining.

Common Mistake

Forgetting the latent heat term. Students compute only the warming heat and end up with a wildly wrong answer. The phase change at 00^\circC is usually the largest single term — never skip it.

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