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

Latent heat of ice is 80 calgm. A man meits 60 gm of ice by chewing in 1 minute. His power is

Heat required to melt 1 g of ice is 80 cal. A man metls 60 g of ice by chewing in one minute. His power is

Latent heat of fusion of ice : 80 cal/g :: Steam :…………..

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 )

10 gm of ice at – 20^(@)C is added to 10 gm of water at 50^(@)C . Specific heat of water = 1 cal//g–.^(@)C , specific heat of ice = 0.5 cal//gm-.^(@)C . Latent heat of ice = 80 cal/gm. Then resulting temperature is -

150 g of ice is mixed with 100 g of water at temperature 80^(@)C . The latent heat of ice is 80 cal/g and the specific heat of water is 1 cal//g-.^(@)C . Assuming no heat loss to the environment, the amount of ice which does not melt is –

A heater melts 0^@C ice in a bucket completely into water in 6 minutes and then evaporates all that water into steam in 47 minutes 30 sec . If latent heat of fusion of ice is 80 cal//gram , latent heat of steam will be (specific heat of water is 1 cal//gam-^@C )