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

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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?

50 g

of ice at

- 5^{o} C

is added to

200 g

of water at

40^{o} C

in a calorimeter. The water equivalent of calorimeter is

50 g

of water. The temperature of mixture is

Determine the final result when 200 gm of water and 20 gm of ice at $0^{\circ} C$ are in a calorimeter having a water equivalent of 30 gm. 50 gm of steam is passed into it at $100^{\circ} C$ . Take latent heat of fusion 80 cal/gm and latent heat of vaporization as 540 cal/gm

Q. A mixture of

250

g of water and

200

g of ice is kept in a calorimeter of water equivalent

50

g. If

200

g of steam at

100^{o}

C is passed through this mixture, calculate the final temperated and weight of contents of the calorimeter. Latent heat of steam is

540

cal

g^{- 1}

Determine the final result when 200 gm of water and 20 gm at $0^{\circ} C$ are in a calorimeter having a water equivalent of 30 gm and 50 gm of steam is passed into it at $100^{\circ} C$ . Take latent heat of fusion 80 cal/gm and latent heat of vaporization as 540 cal/gm

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