Under certain circumstances, both water ice and liquid water can be considered to be incompressible substances with constant specific heats. Determine the change in specific internal energy for
(a) ice as it is cooled from -5° C to -25° C at 1 atm, and
(b) liquid water as it is heated from 20° C to 40°C at 1 atm. For ice, c = 2.04 kJ/kg-K and for liquid water, c = 4.18 kJ/kg-K.
(a) Given: c = 2.04 kJ/kg-K; T1 = -5oC; T2 = -25°C
(b) Given: c = 4.18 kJ/kg-K; T1 = 20oC; T2 = 40°C
What will be an ideal response?
(a) ?u = c (T2 – T1) = -40.8 kJ/kg (Water ice)
(b) ?u = c (T2 – T1) = 83.6 kJ/kg (Liquid water)
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