Home
Class 12
PHYSICS
A calorimeter of water equivalent 20 g...

A calorimeter of water equivalent 20 g contains 180 g of water at `25^(@)C` . 'm' grams of steam at `100^(@)C` is mixed in it till the temperature of the mixture is `31^(@)C` . The value of m is close to (Laten heat of water = 540 `cal g^(-1)` , specific heat of water = 1 cal `g^(-1) ""^(@)C^(-1))`

Promotional Banner

Similar Questions

Explore conceptually related problems

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 )

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 )

60 g of ice at 0^@C is mixed with 60 g of steam at 100^@C . At thermal equilibrium, the mixture contains (Latent heat of steam and ice are 540 g^-1 and 80 cal g^-1 respectively, specific heat of water = 1cal g^-1^@C^-1 )

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 .

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 ).

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