In a laboratory experiment for finding specific latent heat of ice, 100 g of water at 30oC was taken in a calorimeter made of copper and of mass 10 g. When 10 g of ice at 0oC was added to the mixture and kept within the liquid till the ice melted completely, the final temperature of the mixture was found to be 20oC.
Calculate the value of the latent heat of fusion of ice from the data discussed above.
In a laboratory experiment for finding specific latent heat of ice, 100 g of water at 30oC was taken in a calorimeter made of copper and of mass 10 g. When 10 g of ice at 0oC was added to the mixture and kept within the liquid till the ice melted completely, the final temperature of the mixture was found to be 20oC.
Calculate the value of the latent heat of fusion of ice from the data discussed above.
The correct option is D 340 Jg-1
Total quantity of water in calorimeter at 20oC = (100+ 10) g = 110 g.
Heat released by water = 100 x 4.2 x (30-20) = 420 x 10 = 4200 J.
Heat released by calorimeter = 10 x 0.4 x (30-20) = 4 x 10 = 40 J.
The heat gained by ice on melting is expressed as mL + mc ∆t, where m = mass of ice,
L = Latent Heat of fusion, c = specific heat capacity, and ∆t = change in temperature
10L + 10 x 4.2 x (20-0) = 4240
or 10L + 840 = 4240
or 10L = 4240-840 = 3400
or L = 3400/10 = 340 J g-1
Hence, the latent heat of fusion of ice is 340 Jg-1.
340 Jg-1
Total quantity of water in calorimeter at 20oC = (100+ 10) g = 110 g.
Heat released by water = 100 x 4.2 x (30-20) = 420 x 10 = 4200 J.
Heat released by calorimeter = 10 x 0.4 x (30-20) = 4 x 10 = 40 J.
The heat gained by ice on melting is expressed as mL + mc ∆t, where m = mass of ice,
L = Latent Heat of fusion, c = specific heat capacity, and ∆t = change in temperature
10L + 10 x 4.2 x (20-0) = 4240
or 10L + 840 = 4240
or 10L = 4240-840 = 3400
or L = 3400/10 = 340 J g-1
Hence, the latent heat of fusion of ice is 340 Jg-1.
