Earth receives 1400 W//m^2 of solar powe...

Earth receives `1400 W//m^2` of solar power. If all the solar energy falling on a lens of area `0.2m^2` is focused on to a block of ice of mass 280 grams, the time taken to melt the ice will be….. Minutes. (`Latent heat of fusion of ice`=`3.3xx10^5J//kg`.)

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Earth receives 1400 W//m^(2) of solar power. If all the solar energy falling on a lens of area 0.2m^(2) is focussed onto a block of ice of mass 2.80g, it takes 110t(in sec) time to melt the ice. Value of t is:

What is meant by saying that the latent heat of fusion of ice is 3.34 xx 10^(5)J//kg ?

One end of a steel rod ( K =42 J//m-s-^(@)C) of length 1.0m is lept in ice at 0^(@)C and the other end is kept in boiling water at 100^(@)C . The area of cross-section of the rod is 0.04 cm^(2) . Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice = 3.36 xx 10^(5) J//kg .

As 6.00 kg of a liquid substance at its freezing point completely freezes, it gives oft enough heat to melt 3.00 kg of ice at 0^@C . The heat of fusion of the substance [Take latent heat of fusion of ice =3.34xx10^2 kJ//kg ] is

Find the result of mixing 1kg of ice at 0^(@)C with 1.5kg of water at 45^(@)C . (Specific latent heat of fusioin of ice =336xx10^(3)Jkg^(-1) )

How many calories of heat will be absorbed when 2kg of ice at 0^(@)C melts ? ( Specific latent heat of fusion of ice = 80 cal //g )

Find the final temperature and composition of the mixtue of 1 kg of ice at 0^(@)C and 1.5 kg of water at 45^(@)C . Given that specific heat of water is 4200 J/kg and latent heat of fusion of ice is 3.36xx10^(5)J//kg

A container containing Ice at 0°C and having outside surface area (1.6 m^2) and thickness 1 cm is placed in a boiling water if 2 kg of ice melts in 30 minute, then thermal conductivity of material of container is nearly (Given: Latent heat of fusion of ice = 3.36 x 10^5 J/kg)

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