Question

A 9.0 g ice cube at -20 ∘C is in a rigid, sealed container from which all the air has been evacuated. Steam has CV = 1500 J/kg⋅K and CP = 1960 J/kg⋅K. 1-How much heat is required to change this ice cube into steam at 170 ∘C ?

Answer 1

This will be done in following five stages:

1)heat required to get ice from from -20 deg C to 0 deg C

q1 = m C(ice) T1 = 9 x 2.09 x 20 = 376.2 J

2)from 0 deg ice to 0 deg water

q2 = m Hf = 9 x 334 = 3006 J

3)from 0 deg water to 100 deg C

q3 = m C(water) T3 = 9 x 4.18 x 100 = 3762 J

4)from 100 deg C of water to 100 deg C of steam

q4 = m Hv = 9 x 2257 = 20313 J

5)from 100 deg C of steam to 170 deg C
q5 = m c(steam) (170 - 100) = 9 x 2.09 x 70 = 1316.7 J

So the total heat required will be:

Q = q1+q2+q3+q4+q5

Q = 376.2 + 3006 + 3762 + 20313 + 1316.7

Hence, Q = 28773.9 J = 28.77 kJ