Calculate the amount of heat transfer required to change 88 kg of ice (solid state) which is at 00C to
water (liquid state) of 100C. The specific heat capacity of ice is 2090 J/kg0C and the specific latent heat
of fusion is 335 kJ/kg.
1
Expert's answer
2020-05-24T16:33:06-0400
q1 = m x ∆Hfus = 88 kg x 335 kJ/kg = 29 480 kJ
q2 = m x c x ∆T = 88 x 2090 J/kg•°C x (10 - 0)°C = 1 839 kJ
Comments
Leave a comment