Sunrays are allowed to fall on a lens of diameter 20 cm. They are then brought to focus on a calorimeter containing 20 g of ice. If the absorption by the lens is neglible, the time required to melt all the ice is
(solar constant =`1.9 cal min^(-1) cm^(-2) and L = 80 calg^(-1)`)

Rays from the sun ar focuseed by a lens of diameter 5 cm on to a block of ice and 10 g of ice is melted in 20 min. Therefore the heat from the sun reaching the earth per min per square centrimetre is (Latent heat of ice L = 80 cal g^(-1) )

Rays from the sun ar focuseed by a lens of diameter 5 cm on to a block of ice and 10 g of ice is melted in 20 min. Therefore the heat from the sun reaching the earth per min per square centrimetre is (Latent heat of ice L = 80 cal g^(-1) )

Calculate the entropy change when 20.0 g of ice changes to liquid water at 0°C. The heat of fusion is 80.0 cal g^(-1) .

100 gm of water at 20^@C is converted into ice at 0^@C by a refrigerator in 2 hours. What will be the quantity of heat removed per minute? Specific heat of water = 1 cal g^(-1) C^(-1) and latent heat of ice = 80 cal g^(-1)

It takes 20 minutes to melt 10 g of ice, when ray from the sun are focused by a lens of diameter 5 cm on to a block of ice. Calcualate the heat received from the sun of 1cm^(2) per minute. (Given, L = 80 kcal/kg)

About 5g of water at 30^(@)C and 5g of ice at -20^(@)C are mixed together in a calorimeter. Calculate final temperature of the mixture. Water equivalent of the calorimeter is negligible. Specific heat of ice =0.5"cal"g^(-1)C^(@)-1) and latent heat of ice =80"cal"g^(-1)

10g of water at 30^(@)C and 5g of ice at -20^(@) C are mixed together in a calorimeter. What is the final temperature of the mixture? Given specific heat of ice= 0.5 cal g^(-1) (""^(0)C)^(-1) and latent heat of fusion of ice =80 cal g^(-1)

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:

50 g of an alloy containing 80% copper and 20% silver is heated up to t_(1)=80^(@)C and then dropped in a calorimeter of water equivalent W = 10 g containing m = 90 g of water at t_(2)=20^(@)C . What will be the final temerature of the mixture ? Specific heat capacity of copper and silver are s_(c)=0.09" cal.g"^(-1).^(@)C^(-1) and s_(s)="0.15 cal.g"^(-1).^(@)C^(-1) respectively.

500 gms of water is to be cooled from 3.5^@C to 20^@C . How many grams of ice at -10^@C will be required for this purpose? Specific heat of ice = 0.5 cal g^(-1) ""^@C^(-1) Latent heat of ice = 80 cal g^(-1) ?