How much energy is needed to convert 23.0 grams of ice at -10.0°C into steam at 109°C?

Question

Ellie Andrews · Accepted Answer

Converting 23.0 g of ice at -10.0 °C to steam at 109 °C requires 69.8 kJ of energy.

There are five heats to consider:

#q_1# = heat required to warm the ice to 0.00 °C.
#q_2# = heat required to melt the ice to water at 0.00 °C.
#q_3# = heat required to warm the water from 0.00 °C to 100.00 °C.
#q_4# = heat required to vapourize the water to vapour at 100 °C.
#q_5# = heat required to warm the vapour to 109 °C.

#q_1 = mcΔT# = 23.0 g × 2.108 J•°C⁻¹g⁻¹ × 10.0 °C = 484.84 J

#q_2 = mΔH_"fus"# = 23.0 g × 334 J• g⁻¹ = 7682 J

#q_3 = mcΔT# = 23.0 g × 4.184 J°C⁻¹g⁻¹ × 100.00 °C = 9623.2 J

#q_4 = mΔH_"vap"# = 23.0 g × 2260 J•g⁻¹ = 51 980 J

#q_5 = mcΔT# = 23.0 g ×2.01 J•g⁻¹ = 46.23 J

#q_1 + q_2 + q_3 +q_4 +q_5# =( 484.84 + 7682 + 9623.2 + 51 980 + 46.23) J = "69 816.27" J = 69.8 kJ (3 significant figures)

Mateo Aguilar · Answer

undefined To calculate the energy required, use the formula:

$$ Q = m \cdot (c_{\text{ice}} \cdot \Delta T_{\text{ice}} + L_{\text{fusion}} + c_{\text{water}} \cdot \Delta T_{\text{melt}} + L_{\text{vaporization}} + c_{\text{steam}} \cdot \Delta T_{\text{steam}}) $$

$$ Q = 23.0 \, \text{g} \cdot (2.09 \, \text{J/g°C} \cdot (0 - (-10.0°C)) + 334 \, \text{J/g} + 4.18 \, \text{J/g°C} \cdot (100.0°C - 0) + 2260 \, \text{J/g} + 2.01 \, \text{J/g°C} \cdot (109.0°C - 100.0°C)) $$

$$ Q \approx 23.0 \, \text{g} \cdot 748.79 \, \text{J/g} $$

$$ Q \approx 17,213.17 \, \text{J} $$