Determine the amount of heat needed to raise 20.0 g of ice at 0 degrees C to steam at 100 degrees C. (ΔHfusion = 334 J/g; Specific heat of (H2O) = 4.18 J/gc; ΔHvap = 2.25 kJ/g)
To determine the amount of heat needed to raise 20.0 g of ice at 0 degrees Celsius to steam at 100 degrees Celsius, we will break down the process into several steps and calculate the heat absorbed or released at each step.
Step 1: Heating the ice from -273 to 0 degrees Celsius.
q1 = mass × specific heat × ΔT
q1 = 20.0 g × 2.09 J/g°C × (0 - (-273) °C)
q1 = 20.0 g × 2.09 J/g°C × 273 °C
q1 = 11,353.4 J or 11.3534 kJ
Step 2: Melting the ice at 0 degrees Celsius.
q2 = mass × ΔHfusion
q2 = 20.0 g × 334 J/g
q2 = 6,680 J or 6.68 kJ
Step 3: Heating the water from 0 to 100 degrees Celsius.
q3 = mass × specific heat × ΔT
q3 = 20.0 g × 4.18 J/g°C × (100 - 0) °C
q3 = 8,360 J or 8.36 kJ
Step 4: Vaporizing the water at 100 degrees Celsius.
q4 = mass × ΔHvap
q4 = 20.0 g × 2.25 kJ/g
q4 = 45 kJ
The total heat required is the sum of the heat absorbed in each step:
Total heat = q1 + q2 + q3 + q4
Total heat = 11.3534 kJ + 6.68 kJ + 8.36 kJ + 45 kJ
Total heat = 71.3934 kJ
Therefore, the amount of heat needed to raise 20.0 g of ice at 0 degrees Celsius to steam at 100 degrees Celsius is 71.3934 kJ.