A calorimeter (of water equivalent 50 g ) contains 250 g of water and 50 g of ice at 0 ∘ C . 30 g of water at 8 0 ∘ C is added to it. The final condition of the system will be [ s w a t er = 1 cal/g

Question

A calorimeter (of water equivalent 50 g ) contains 250 g of water and 50 g of ice at 0 ∘ C . 30 g of water at 8 0 ∘ C is added to it. The final condition of the system will be [ s w a t er ​ = 1 cal/g- ∘ C , L f ​ = 80 cal/g]

TestHub · Accepted Answer

Heat available from the hot water cooling to $0^{\circ} C$ :
$Δ Q_{a v} = 30 \times 1 \times 80 = 2400 cal$
Heat needed to melt all the ice:
$Δ Q_{re q} = 50 \times 80 = 4000 cal$
$2400 < 4000$
Melting is not complete, so the temperature stays $0^{\circ} C$ . Mass of ice melted:
$m \times 80 = 2400$
$m = 30 g$
$m_{i ce} = 50 - 30 = 20 g$

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