Question

Calculate the total amount of heat energy in kilocalories that must be released in order to condense 18.9 g of steam at 373.0 K, cool that liquid to 272.0 K, then freeze the liquid at 272.0 K.

Answer 1

Answer on Question #81410, Chemistry/ Organic Chemistry

Calculate the total amount of heat energy in kilocalories that must be released in order to condense $18.9\mathrm{g}$ of steam at $373.0\mathrm{K}$, cool that liquid to $272.0\mathrm{K}$, then freeze the liquid at $272.0\mathrm{K}$.

Solution

latent heat of fusion of ice $= 335\mathrm{J / g}$

latent heat of vaporisation of water $= 2260\mathrm{J / g}$

specific heat capacity of ice $= 2.14\mathrm{J / (g^{\circ}C)}$

specific heat capacity of water $= 4.184\mathrm{J / (g^{\circ}C)}$

specific heat capacity of steam $= 2.01\mathrm{J / (g^{\circ}C)}$

The energy released is determined in 4 stages:

$$Q = Q _ {1} + Q _ {2} + Q _ {3} + Q _ {4}$$

1. Find $Q_{1}$ — the heat of phase change (steam-water) at 373.0 K (or $T(^{\circ}C) = 373 - 273 = 100^{\circ}C$):

$Q_{1} = L_{v}\times m$, where $L_{v} = 2260\mathrm{J / g}$

$Q_{1} = 2260\mathrm{J / g}\times 18.9\mathrm{g} = 2.23\cdot 10^{6}\mathrm{J} = 42714\mathrm{J}$

As Q is released then $Q_{1} = -42714\mathrm{J}$

2. Find $Q_{2}$ — heat that is released when the temperature of water changes from 373 K to 273 K (or from $100^{\circ}C$ to $0^{\circ}C$)

$Q_{2} = \mathrm{cm}(T_{2} - T_{1})$, where c (for water) $= 4.184\mathrm{J / g\cdot^{\circ}C}$

$Q_{3} = 4.184\mathrm{J / g\cdot^{\circ}C}\cdot 18.9\mathrm{g}\cdot (0 - 100)^{\circ}\mathrm{C} = -7907.8\mathrm{J}$

3. Find $Q_{3}$ — the heat of phase change (water-ice):

$Q_{3} = L_{f}\times m$, where $L_{f} = 335\mathrm{J / g}$

$Q_{3} = 335\mathrm{J / g}\cdot 18.9\mathrm{g} = 6331.5\mathrm{J}$

As Q is released then $Q_{3} = -6331.5\mathrm{J}$

4. Find $Q_{4}$ — heat that is released when the temperature changes from 273 K to 272 K (or from $0^{\circ}C$ to $-1^{\circ}C$)

$Q_{4} = \mathrm{cm}(T_{2} - T_{1})$, where c (for ice) $= 2.14\mathrm{J / g\cdot^{\circ}C}$

$Q_{4} = 2.14\mathrm{J / g\cdot^{\circ}C}\cdot 18.9\mathrm{g}\cdot (-1 - 0)^{\circ}\mathrm{C} = -40.4\mathrm{J}$

$$Q = Q _ {1} + Q _ {2} + Q _ {3} + Q _ {4} = - 4 2 7 1 4 J - 7 9 0 7. 8 J - 6 3 3 1. 5 J - 4 0. 4 J = - 5 6 9 9 3. 7 J$$

The sign “-” shows that the system loses 56993.7 J of energy releasing this amount of energy into surrounding. So, surrounding gets is 56993.7 J of energy or $56993.7\mathrm{J}\times 0.2388\mathrm{cal / J} = 13610\mathrm{cal} = 13.6\mathrm{kcal}$.

Answer: 13.6 kcal

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