Question

Ice water mixture is in equilibrium in a
rigid container. The amount of heat
supplied to decrease the volume of
mixture by 1 cm^3 without any change in
temperature is
equal to
(Pice = 0.9g/cc and L of ice = 80cal/g)
1) 1000 cal
2) 720 cal
3) 500cal
d) 360 cal
​

Answer 1

Answer:

Explanation:x gm ice convert into x gm water

x0.9−x=1⇒x=0.90.1=9

∴Q=9×80=720cal.