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

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

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Answer 1 · 2014-07-25T20:16:19

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:

$q_{1}$ $q_1$ = heat required to warm the ice to 0.00 °C.
$q_{2}$ $q_2$ = heat required to melt the ice to water at 0.00 °C.
$q_{3}$ $q_3$ = heat required to warm the water from 0.00 °C to 100.00 °C.
$q_{4}$ $q_4$ = heat required to vapourize the water to vapour at 100 °C.
$q_{5}$ $q_5$ = heat required to warm the vapour to 109 °C.

$q_{1} = m c Δ T$ $q_1 = mcΔT$ = 23.0 g × 2.108 J•°C⁻¹g⁻¹ × 10.0 °C = 484.84 J

$q_{2} = m Δ H_{fus}$ $q_2 = mΔH_"fus"$ = 23.0 g × 334 J• g⁻¹ = 7682 J

$q_{3} = m c Δ T$ $q_3 = mcΔT$ = 23.0 g × 4.184 J°C⁻¹g⁻¹ × 100.00 °C = 9623.2 J

$q_{4} = m Δ H_{vap}$ $q_4 = mΔH_"vap"$ = 23.0 g × 2260 J•g⁻¹ = 51 980 J

$q_{5} = m c Δ T$ $q_5 = mcΔT$ = 23.0 g ×2.01 J•g⁻¹ = 46.23 J

$q_{1} + q_{2} + q_{3} + q_{4} + q_{5}$ $q_1 + q_2 + q_3 +q_4 +q_5$ =( 484.84 + 7682 + 9623.2 + 51 980 + 46.23) J = "69 816.27" J = 69.8 kJ (3 significant figures)

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

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