How many gm of ice at − 2 0 ∘ C are needed to cool 200 gm of water from 2 5 ∘ C to 1 0 ∘ C ?

Heat lost by water: Q = 200 × 1 × ( 25 − 10 ) = 3000 cal Heat gained by ice of mass m (warm to 0 ∘ C , melt, warm to 1 0 ∘ C ): Q = m × 2 1 × 20 + m × 80 + m × 1 × 10 Q = 100 m Heat lost = heat gained: 3000 = 100 m m = 30 gm

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PhysicsHeat & ThermoCalorimetryMedium2 minQB

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How many gm of ice at $- 2 0^{\circ} C$ are needed to cool $200 gm$ of water from $2 5^{\circ} C$ to $1 0^{\circ} C$ ?

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Answer:

Solution:

Heat lost by water:
$Q = 200 \times 1 \times (25 - 10) = 3000 cal$
Heat gained by ice of mass $m$ (warm to $0^{\circ} C$ , melt, warm to $1 0^{\circ} C$ ):
$Q = m \times \frac{1}{2} \times 20 + m \times 80 + m \times 1 \times 10$
$Q = 100 m$
Heat lost = heat gained:
$3000 = 100 m$
$m = 30 gm$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Calorimetry

⏱ 2mℹ️ Source: QB

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