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A calorimeter of mass 100 g (specific heat 0.1 cal/g °C) contains 250 g of liquid at 30 °C (specific heat 0.4 cal/g °C). A 10 g piece of ice at 0 °C is dropped in. Using L(fusion) = 80 cal/g and specific heat of water = 1 cal/g °C, the final temperature is closest to:
A30.0 °C
B0 °C
C25.2 °C
D20.8 °C
Answer & Solution
Correct answer: D. 20.8 °C
1. Heat available from calorimeter + liquid cooling from 30 °C to T: $(100 \times 0.1 + 250 \times 0.4)(30 - T) = 110(30 - T)$.
2. Heat used by ice = melt + warm melt-water 0 °C to T: $10 \times 80 + 10 \times 1 \times (T - 0) = 800 + 10T$.
3. Heat exchange: $110(30 - T) = 800 + 10T$.
4. Expand: $3300 - 110T = 800 + 10T$, so $120T = 2500$.
5. $T = 2500 / 120 \approx 20.8\ ^\circ C$.
6. Forgetting the latent heat of the ice raises T above the true value, so the 25.2 °C trap is rejected.
_Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 5 "Heat", p.81_
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