A88-kg sample of liquid water is cooled from 0°c to it freezes in the process. how much heat is liberated? for water and the specific heat capacity of ice is .

During a change of state from liquid to solid (like in this problem, from liquid water to ice), the amount of heat that it is liberated is given by

where
m is the mass of the substance
is the latent heat of fusion of the substance.

For ice, the latent heat of fusion is , while the mass of the water in this problem is m=88 kg. If we substitute these data into the equation, we find the amount of heat liberated:

The answer is orbit around nucleus

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward