Question

How much heat is needed to transform 59.9 grams of Ice at-75 C to steam at 100

Answer 1

mcΔT= 59.9g × 2.108 J•°C⁻¹g⁻¹ × 75 °C = 9470 J

mcΔH_fus = 59.9g × 334 J• g⁻¹ = 20007J

mcΔT = 59.9g × 4.184 J°C⁻¹g⁻¹ × 100.00 °C = 25062J

mcΔH_vap = 59.9g × 2260 J•g⁻¹ = 135374J

total heat = 135374J +25062J + 20007J + 9470 J

= 189913 J