A piece of ice (heat capacity=`2100Jkg^-1^@C^-1` and latent heat `=3.36xx10^5Jkg^-1`) of mass m grams is at `-5^@C` at atmospheric pressure. It is given 420J of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that 1gm of ice has melted. Assuming there is no other heat exhange in the process, the value of m is

A piece of ice (heat capacity =2100 J kg^(-1) .^(@)C^(-1) and latent heat =3.36xx10^(5) J kg^(-1) ) of mass m grams is at -5 .^(@)C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice . Water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m in gram is

A piece of ice (heat capacity =2100 J kg^(-1) .^(@)C^(-1) and latent heat =3.36xx10^(5) J kg^(-1) ) of mass m grams is at -5 .^(@)C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice . Water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m in gram is

What does it mean to say that latent heat of ice equals 3.36xx10^5Jkg^(-1) ?

One gram of ice at 0^(@)C is added to 5 gram of water at 10^(@)C . If the latent heat of ice be 80 cal/g, then the final temperature of the mixture is -

A cube of iron (density = 8000kg m^(-1) , specific heat capacity = 470g J kg^(-1)K^(-1) ) is heated to a high temperature and is placed on a larger block of ice at 0^(@)C . The cube melts the ice below it, displaces the water and sinks in the final equilibrium position its upper surface is just inside the ice. Calculate the initial temperature of the cube. Neglect any loss of heat outside the ice and the cube .The density of ice = 900 kg m^(-1) and the latent heat of fusion of ice = 3.36 xx 10^(5)J kg^(-1)

Latent heat of ice 80 cal/gm . A man melts 60 g of ice by chewing in 1 minute . His power is