Question

1. how much energy (heat) is required to convert 52.0 g of ice at -10.0°c to steam at 100°c?

specific heat of ice: 2.09 j/g·°c δh_fus = 6.02 kj/mol
specific heat of water: 4.18 j/g·°c δh_vap = 40.7 kj/mol
specific heat of steam: 1.84 j/g·°c

40.2 kj
22.95 kj
25.70 kj
157.8 kj
1.09 kj

Explanation:

Step1: Heat ice to 0°C

$q_1 = m \cdot c_{ice} \cdot \Delta T = 52.0g \cdot 2.09J/g^\circ C \cdot (0 - (-10))^\circ C = 52.0 \cdot 2.09 \cdot 10 = 1086.8J = 1.0868kJ$

Step2: Melt ice to water

Moles of $H_2O = \frac{52.0g}{18.02g/mol} \approx 2.886mol$
$q_2 = n \cdot \Delta H_{fus} = 2.886mol \cdot 6.02kJ/mol \approx 17.37kJ$

Step3: Heat water to 100°C

$q_3 = m \cdot c_{water} \cdot \Delta T = 52.0g \cdot 4.18J/g^\circ C \cdot (100 - 0)^\circ C = 52.0 \cdot 4.18 \cdot 100 = 21736J = 21.736kJ$

Step4: Vaporize water to steam

$q_4 = n \cdot \Delta H_{vap} = 2.886mol \cdot 40.7kJ/mol \approx 117.46kJ$

Step5: Total heat

$Q = q_1 + q_2 + q_3 + q_4 \approx 1.0868 + 17.37 + 21.736 + 117.46 \approx 157.65kJ$

Answer:

157.8 kJ (closest option)