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20g stream at 100°C is passed through 15...

20g stream at 100°C is passed through 15g ice at -12°C. The mass of water in in calorie metre will be nearly (specific heat of ice =0.5 cal/g°C, specific heat of water=1 cal/g°C latent heat of vaporisation 540 cal/g, latent heat of fusion=80 cal/g and neglect heat capacity of calorimeter )

A

20g water at 100°C

B

25g water at 100°C

C

18g water at 100°C

D

15g water at 100°C

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2kg ice at -20"^(@)C is mixed with 5kg water at 20"^(@)C . Then final amount of water in the mixture will be: [specific heat of ice =0.5 cal//gm "^(@)C , Specific heat of water =1 cal//gm"^(@)C , Latent heat of fusion of ice = 80 cal//gm]

If 10 g of ice at 0^(@)C is mixed with 10 g of water at 40^(@)C . The final mass of water in mixture is (Latent heat of fusion of ice = 80 cel/g, specific heat of water =1 cal/g""^(@)C )

20 g of steam at 100^@ C is passes into 100 g of ice at 0^@C . Find the resultant temperature if latent heat if steam is 540 cal//g ,latent heat of ice is 80 cal//g and specific heat of water is 1 cal//g^@C .

The water equivatent of a 400 g copper calorimeter (specific heat =0.1 "cal"//g^(@)C )

Specific latent heat of fusion is expressed in g/cal.

1 g of steam at 100^@C and an equal mass of ice at 0^@C are mixed. The temperature of the mixture in steady state will be (latent heat of steam =540 cal//g , latent heat of ice =80 cal//g ,specific heat of water =1 cal//g^@C )

The amount of heat required to raise the temperature of 75 kg of ice at 0°C to water at 10°C is (latent heat of fusion of ice is 80 cal/g, specific heat of water is 1 cal/g °C)

50 g of steam at 100^@ C is passed into 250 g of at 0^@ C . Find the resultant temperature (if latent heat of steam is 540 cal//g , latent heat of ice is 80 cal//g and specific heat of water is 1 cal//g -.^@ C ).

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