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At 1 atm, how much energy is required to heat 55.0 g H2O(s) at -24.0 °C to H,O(g) at 139.0 °C? Use the heat transfer constants found in this table.

2 Answers

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Final answer:

To calculate the total energy required to heat 55.0 g of ice at -24.0 °C to steam at 139.0 °C, the energies for each step of heating solid, phase change, heating liquid, another phase change, and heating gas are calculated and summed, using specific heats and heats of fusion and vaporization.

Step-by-step explanation:

Calculating Energy Required for Heating Water

The process of heating 55.0 g of H2O(s) from -24.0 °C to H2O(g) at 139.0 °C involves several steps, each requiring a different amount of energy. These energy changes include heating the ice to 0 °C, melting the ice, heating the resulting water to 100 °C, and then vaporizing the water into steam. Lastly, we must heat the steam from 100 °C to 139 °C.The specific heats of ice, water, and steam, along with the heats of fusion and vaporization, are necessary to calculate the total energy. The energy required for each step can be calculated using the formula q = mcΔT, where q is the energy in Joules, m is the mass in grams, c is the specific heat capacity, and ΔT is the temperature change. The heats of fusion and vaporization are used when the phase changes from solid to liquid and from liquid to gas, respectively.

To find the total heat required, sum the energies calculated for each separate process: heating ice, melting ice, heating water, vaporizing water, and heating steam. To solve the problem, knowledge of thermodynamics, specific heat, heat of fusion, and heat of vaporization is applied, all of which fall under the domain of chemistry and are typically taught at the high school level.

User Mzdv
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8.7k points
5 votes

A total of 1342218.4 J of energy is required to heat 55.0 g of ice at -24.0 °C to steam at 139.0 °C at 1 atm.

Heating of ice from -24.0 °C to 0 °C

energy required Q_ice = m_ice * c_ice * ΔT_ice Q_ice

= 55.0 g * 2.093 J/g°C * (0 °C - (-24.0 °C))

= 2823.4 J

Melting ice at 0 °C:

energy required Q_fusion = m_ice * ΔH_fusion_ice Q_fusion

= 55.0 g * 333.55 J/g Q_fusion

=18345.8 J

Heating of water from 0 °C to 139.0 °C:

energy required : Q_water = m_water * c_water * ΔT_water Q_water

= 55.0 g * 4.184 J/g°C * (139.0 °C - 0 °C)

= 29709.6 J

Vaporizing of water at 139.0 °C:

energy required: Q_vaporization = n_water * ΔH_vaporization_water * 1000 J/kJ Q_vaporization

= 1240540 J

Therefore the total energy required:

Q_total = Q_ice + Q_fusion + Q_water + Q_vaporization

Q_total = 1342218.4 J

User ULick
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