This worked example problem demonstrates how to calculate the energy required to raise the temperature of a sample that includes changes in phase. This problem finds the energy required to turn cold ice into hot steam. What is the heat in Joules required to convert 25 grams of -10 °C ice into 150 °C steam?Useful information:heat of fusion of water = 334 J/gheat of vaporization of water = 2257 J/gspecific heat of ice = 2.09 J/g·°Cspecific heat of water = 4.18 J/g·°Cspecific heat of steam = 2.09 J/g·°C The total energy required is the sum of the energy to heat the -10 °C ice to 0 °C ice, melting the 0 °C ice into 0 °C water, heating the water to 100 °C, converting 100 °C water to 100 °C steam and heating the steam to 150 °C. To get the final value, first calculate the individual energy values and then add them up. Step 1: Find the heat required to raise the temperature of ice from -10 °C to 0 °C. Use the formula: q = mcΔT where
In this problem:
Plug in the values and solve for q: q = (25 g)x(2.09 J/g·°C)[(0 °C - -10 °C)]q = (25 g)x(2.09 J/g·°C)x(10 °C) q = 522.5 J
Find the heat required to convert 0 °C ice to 0 °C water.
q = m·ΔHf where
For this problem:
Plugging in the values gives the value for q:
q = (25 g)x(334 J/g) The heat required to convert 0 °C ice to 0 °C water = 8350 J
Find the heat required to raise the temperature of 0 °C water to 100 °C water.q = mcΔTq = (25 g)x(4.18 J/g·°C)[(100 °C - 0 °C)]q = (25 g)x(4.18 J/g·°C)x(100 °C)q = 10450 JThe heat required to raise the temperature of 0 °C water to 100 °C water = 10450 J Step 4:
Find the heat required to convert 100 °C water to 100 °C steam. q = heat energy m = massΔHv = heat of vaporization q = (25 g)x(2257 J/g)q = 56425 JThe heat required to convert 100 °C water to 100 °C steam = 56425 Step 5: Find the heat required to convert 100 °C steam to 150 °C steamq = mcΔTq = (25 g)x(2.09 J/g·°C)[(150 °C - 100 °C)]q = (25 g)x(2.09 J/g·°C)x(50 °C)q = 2612.5 J The heat required to convert 100 °C steam to 150 °C steam = 2612.5 Step 6: Find total heat energy. In this final step, put together all of the answers from the previous calculations to cover the entire temperature range.
Answer: The heat required to convert 25 grams of -10 °C ice into 150 °C steam is 78360 J or 78.36 kJ.
Phase changes When a system receives an amount of energy q at a constant temperature, T, the entropy increase DS is defined by the following equation. Hence, the magnitude of ΔS for a reversible process such as a phase change is calculated \[ \Delta S = \dfrac{q_{rev}}{T}\] with the temperature in Kelvin. Since entropy changes are much smaller than enthalpy changes, they are usually reported in J K–1 mol–1.
Examples of reversible processes are For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. For example, combustion of a fuel in air involves transfer of heat from a system (the fuel and oxygen molecules undergoing reaction) to surroundings that are infinitely more massive (the earth’s atmosphere). As a result, \(q_{surr}\) is a good approximation of \(q_{rev}\), and the second law may be stated as the following: \[ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \label{4}\] We may use this equation to predict the spontaneity of a process as illustrated in Example \(\PageIndex{1}\).
The entropy change for the process \[\ce{H2O}(s)⟶\ce{H2O}(l) \nonumber\] is 22.1 J/K and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at −10.00 °C? Is it spontaneous at +10.00 °C?
We can assess the spontaneity of the process by calculating the entropy change of the universe. If ΔSuniv is positive, then the process is spontaneous. At both temperatures, ΔSsys = 22.1 J/K and qsurr = −6.00 kJ. At −10.00 °C (263.15 K), the following is true: \[\begin{align} ΔS_\ce{univ}&=ΔS_\ce{sys}+ΔS_\ce{surr}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \nonumber\\ &=\mathrm{22.1\: J/K+\dfrac{−6.00×10^3\:J}{263.15\: K}=−0.7\:J/K} \nonumber \end{align} \nonumber\] \(S_{univ} < 0\), so melting is nonspontaneous (not spontaneous) at −10.0 °C. At 10.00 °C (283.15 K), the following is true: \[ΔS_\ce{univ}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T}\nonumber\] \[\mathrm{=22.1\:J/K+\dfrac{−6.00×10^3\:J}{283.15\: K}=+0.9\: J/K}\nonumber\] \(S_{univ} > 0\), so melting is spontaneous at 10.00 °C.
Using this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of Suniv? Answer: Entropy is a state function, and freezing is the opposite of melting. At −10.00 °C spontaneous, +0.7 J/K; at +10.00 °C nonspontaneous, −0.9 J/K.
Ice melting provides an example in which entropy increases in a small system, a thermodynamic system consisting of the surroundings (the warm room) and the entity of glass container, ice, water which has been allowed to reach thermodynamic equilibrium at the melting temperature of ice. In this system, some heat (δQ) from the warmer surroundings at 298 K (25 °C; 77 °F) transfers to the cooler system of ice and water at its constant temperature (T) of 273 K (0 °C; 32 °F), the melting temperature of ice. The entropy of the system, which is δQ/T, increases by δQ/273K. The heat δQ for this process is the energy required to change water from the solid state to the liquid state, and is called the enthalpy of fusion, i.e. ΔH for ice fusion. It is important to realize that the entropy of the surrounding room decreases less than the entropy of the ice and water increases: the room temperature of 298 K is larger than 273 K and therefore the ratio, (entropy change), of δQ/298K for the surroundings is smaller than the ratio (entropy change), of δQ/273K for the ice and water system. This is always true in spontaneous events in a thermodynamic system and it shows the predictive importance of entropy: the final net entropy after such an event is always greater than was the initial entropy. As the temperature of the cool water rises to that of the room and the room further cools imperceptibly, the sum of the δQ/T over the continuous range, “at many increments”, in the initially cool to finally warm water can be found by calculus. The entire miniature ‘universe’, i.e. this thermodynamic system, has increased in entropy. Energy has spontaneously become more dispersed and spread out in that ‘universe’ than when the glass of ice and water was introduced and became a 'system' within it.
Calculate entropy change when 36.0 g of ice melts at 273 K and 1 atm. If you look up the enthalpy of fusion for ice in a table, you would get a molar enthalpy of 6.01 kJ/mol.
dS = (6.01 kJ mol-1)/272 K * (36 g)/(18 g mol-1) = 1.22 kJ / K Contributors and Attributions |
What is the change in entropy of 1,000 kg of ice at 0.0 when it melts and becomes steam at 100.0 c
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