A volume of 120 ml of drink (half alcohol + half water by mass) originally at a temperature of 25 ∘ C is cooled by adding 20 gm ice at 0 ∘ C . If all the ice melts, the final temperature of the drink is : (density of drink = 0.833 gm /cc, specific heat of alcohol = 0.6 cal/gm / ∘ C )

m = 0.833 ( 120 ) m ≈ 100 g Equal masses of water and alcohol give heat capacity: C = 50 ( 1 ) + 50 ( 0.6 ) C = 80 cal/ ∘ C Heat lost by the drink melts and warms the ice. 80 ( 25 − T ) = 1600 + 20 T 400 = 100 T T ≈ 4 ∘ C

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PhysicsHeat & ThermoCalorimetryMedium2 minQB

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A volume of $120 ml$ of drink (half alcohol + half water by mass) originally at a temperature of $25^{\circ} C$ is cooled by adding $20 gm$ ice at $0^{\circ} C$ . If all the ice melts, the final temperature of the drink is : (density of drink = $0.833 gm$ /cc, specific heat of alcohol = $0.6 cal/gm$ / $^{\circ} C$ )

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Answer:

Solution:

$m = 0.833 (120)$
$m \approx 100 g$
Equal masses of water and alcohol give heat capacity:
$C = 50 (1) + 50 (0.6)$
$C = 80 cal/^{\circ} C$
Heat lost by the drink melts and warms the ice.
$80 (25 - T) = 1600 + 20 T$
$400 = 100 T$
$T \approx 4^{\circ} C$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Calorimetry

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