In a calorimeter water equivalent - 40 gm are 200 gm of water and 50 gm of ice all at 0- C- Into this is poured 30 gm of water at 90- C- What will be the final condition of the system-A- 280 gm ice at 0- CB- 280 gm water at 0- CC- 280 gm water at 4- CD- 16-25 gm ice- 266-75 gm water

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Boyle's Law

In a calorime...

Question

In a calorimeter (water equivalent = 40 gm) are 200 gm of water and 50 gm of ice all at $0^{\circ} C$ . Into this is poured 30 gm of water at $90^{\circ} C$ . What will be the final condition of the system?

280 gm ice at 0°C

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280 gm water at 0°C

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280 gm water at 4°C

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16.25 gm ice, 266.75 gm water

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Solution

The correct option is D
16.25 gm ice, 266.75 gm water

Let us assume that all ice melts and warms up. Thus we will assume that $T > 0$ .

Heat lost by water added = Heat gained by ice to melt

Heat to warm water formed from ice and water added

Heat gained by calorimeter can.

$\Rightarrow$ $30 \times 1 \times (90 - T)$ = $50 \times 80 + (50 + 200) \times 1 \times (T - 0) + 40 \times 1 \times (T - 0)$

$\Rightarrow$ $2700 - 30 T$ = $4000 + 250 T + 40 T$

$\Rightarrow$ $T$ = $- {4.1}^{\circ} C$

Hence our assumption that $T > 0$ is wrong, since hot water added is not able to melt all of the ice.

Therefore the final temperature will be $0^{\circ} C$

Let $m$ = mass of ice finally left in the can.

Heat lost by water = heat gained by melting ice

$\Rightarrow$ $30 \times 1 \times (90 - 0)$ = $(50 - m) \times 80$

$\Rightarrow$ $m$ = 16.25 gm

Finally there is 16.25 gm of ice and (200 + 30 + 33.75) = 266.75 gm of water at $0^{\circ} C$

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In a calorimeter (water equivalent = 40 gm) are 200 gm of water and 50 gm of ice all at 0∘C. Into this is poured 30 gm of water at 90∘C. What will be the final condition of the system?

In a calorimeter (water equivalent = 40 gm) are 200 gm of water and 50 gm of ice all at $0^{\circ} C$ . Into this is poured 30 gm of water at $90^{\circ} C$ . What will be the final condition of the system?