Question

3.6 kilograms of water are changed (a) from ice at 0?C into liquid water at 0?C and (b) from liquid water at 100?C into steam at 100?C. For each situation, determine the change in mass of the water.

Answer 1

Specific latent heat=heat required to change the substance/(mass of substance)

l=Q/M

a)

This requires a change of state solid to liquid

Q    =   m l_f

Q    = 3.6x 3.35 x 10⁵    = 1206000 J (latent heat of water at melting point is 3.35*10^5)

change in mass=Q/c^2=1206000/(3*10^8)^2=13.4*10^-12 0r 1.34*10^-11

b)

This is a change of state liquid to gas

Q    =    m l_v

Q    = 3.6 x 2.26 x 10⁶ = 8136000 J(latent heat of vaparisation =2.26*10^6)

change in mass=Q/c^2=9.04*10^-11