Engineering Board Exam Reviewer · Thermodynamics · hard
How much heat is required to convert 2 kg of ice at −10 °C into water at 20 °C? (cice = 2.09 kJ/kg· K, h if = 335 kJ/kg, cwater = 4.187 kJ/kg· K)
- 837.48 kJ
- 209.28 kJ
- 921.22 kJ
- 879.28 kJ
Answer
D. 879.28 kJ
Solution
Q = m cice T + m h if + m cw T = 41.80 + 670.00 + 167.48 = 879.28 kJ.
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