10 gm of ice at 0 ∘ C is mixed with 'm' gm of water at 50 ∘ C . What is minimum value of m so that ice melts completely. ( L f = 80 cal/gm and s w = 1 cal/gm - ∘ C )

For minimum water mass, the final temperature is 0 ∘ C . Q ice = 10 ( 80 ) = 800 cal Q water = m ( 1 ) ( 50 ) 800 = 50 m m = 16 g

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$10 gm$ of ice at $0^{\circ} C$ is mixed with 'm' gm of water at $50^{\circ} C$ . What is minimum value of m so that ice melts completely. ( $L_{f}$ = $80 cal/gm$ and $s_{w}$ = $1 cal/gm$ - $^{\circ} C$ )

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Answer:

Solution:

For minimum water mass, the final temperature is $0^{\circ} C$ .
$Q_{ice} = 10 (80) = 800 cal$
$Q_{water} = m (1) (50)$
$800 = 50 m$
$m = 16 g$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Calorimetry

⏱ 2mℹ️ Source: QB

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