find the result of mixing 10 g of ice at -10 degree C with 10 g of water at 10 degree C. specific heat capacity of ice = 2.1 J g-1 K-1, specific latent heat of ice = 336 J g-1 and specific heat capacity of water = 4.2 J g-1 K-1.
find the result of mixing 10 g of ice at -10 degree C with 10 g of water at 10 degree C. specific heat capacity of ice = 2.1 J g-1 K-1, specific latent heat of ice = 336 J g-1 and specific heat capacity of water = 4.2 J g-1 K-1.
Specific heat of water is = 4.2 J/(g.K)
Specific heat of ice is = 2.1 J/(g.K)
Latent heat of ice is = 336 J/g
Heat released by 10 g of water at 10 oC in converting to water at 0 oC is,
H1 = msθ = (10)(4.2)(10) = 420 J
Heat gained by 10 g of ice at -10 oC in converting into ice at 0 oC is,
H2 = (10)(2.1)(10) = 210 J
Suppose, ‘x’ g of ice now melts. So, heat gained by this ‘x’ g of ice in converting into water at 0 oC is,
H3 = xL = 336x J
Now, H3 + H2 = H1
=> 336x + 210 = 420
=> x = 0.625 g
Mass of ice remaining is = 10 – 0.625 = 9.375 g
Thus, in the resulting mixture the ice remaining is 9.375 g and the temperature of the mixture is 0 oC.
Specific heat of water is = 4.2 J/(g.K)
Specific heat of ice is = 2.1 J/(g.K)
Latent heat of ice is = 336 J/g
Heat released by 10 g of water at 10 oC in converting to water at 0 oC is,
H1 = msθ = (10)(4.2)(10) = 420 J
Heat gained by 10 g of ice at -10 oC in converting into ice at 0 oC is,
H2 = (10)(2.1)(10) = 210 J
Suppose, ‘x’ g of ice now melts. So, heat gained by this ‘x’ g of ice in converting into water at 0 oC is,
H3 = xL = 336x J
Now, H3 + H2 = H1
=> 336x + 210 = 420
=> x = 0.625 g
Mass of ice remaining is = 10 – 0.625 = 9.375 g
Thus, in the resulting mixture the ice remaining is 9.375 g and the temperature of the mixture is 0 oC.
