How many minimum ice cubes of mass 10 gm at 0 ∘ C each should added to a glass containing 100 gm water at 2 0 ∘ C to bring it to 0 ∘ C ?

Heat given by water cooling to 0 ∘ C equals heat to melt the ice: m × 80 = 100 × 1 × ( 20 − 0 ) m = 25 gm n = 10 25 = 2.5 Cubes are whole, so the minimum is 3.

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PhysicsHeat & ThermoCalorimetryMedium2 minQB

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How many minimum ice cubes of mass $10 gm$ at $0^{\circ} C$ each should added to a glass containing $100 gm$ water at $2 0^{\circ} C$ to bring it to $0^{\circ} C$ ?

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Answer:

Solution:

Heat given by water cooling to $0^{\circ} C$ equals heat to melt the ice:
$m \times 80 = 100 \times 1 \times (20 - 0)$
$m = 25 gm$
$n = \frac{25}{10} = 2.5$
Cubes are whole, so the minimum is 3.

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Calorimetry

⏱ 2mℹ️ Source: QB

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