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Chapter 6: Problem 131

You have a 1.00-mole sample of water at \(-30 .{ }^{\circ} \mathrm{C}\) and you heat it until you have gaseous water at \(140 .{ }^{\circ} \mathrm{C}\). Calculate \(q\) for the entire process. Use the following data. Specific heat capacity of ice \(=2.03 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) Specific heat capacity of water \(=4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) Specific heat capacity of steam \(=2.02 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) \(\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H_{\text {fusion }}=6.02 \mathrm{~kJ} / \mathrm{mol}\left(\right.\) at \(\left.0^{\circ} \mathrm{C}\right)\) \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H_{\text {vaporization }}=40.7 \mathrm{~kJ} / \mathrm{mol}\left(\right.\) at \(\left.100 .^{\circ} \mathrm{C}\right)\)

Short Answer

Expert verified
The total heat, \(q_{\text{total}}\), required for the entire process is calculated by summing up the heats from all stages: heating the ice, melting the ice, heating the water, vaporizing the water, and heating the steam. The calculations are as follows: 1. Heating the ice: \(q_{\text{heating ice}} = 18.0 \ \text{g} \times 2.03 \ \frac{\text{J}}{^\circ \text{C} \cdot \text{g}} \times 30 \ ^\circ\text{C} = 1096.2 \ \text{J}\) 2. Melting the ice: \(q_{\text{melting ice}} = 1.00 \ \text{mol} \times 6.02 \ \frac{\text{kJ}}{\text{mol}} \times \frac{1000 \ \text{J}}{\text{kJ}} = 6020 \ \text{J}\) 3. Heating the water: \(q_{\text{heating water}} = 18.0 \ \text{g} \times 4.18 \ \frac{\text{J}}{^\circ \text{C} \cdot \text{g}} \times 100 \ ^\circ\text{C} = 7532.4 \ \text{J}\) 4. Vaporizing the water: \(q_{\text{vaporizing water}} = 1.00 \ \text{mol} \times 40.7 \ \frac{\text{kJ}}{\text{mol}} \times \frac{1000 \ \text{J}}{\text{kJ}} = 40700 \ \text{J}\) 5. Heating the steam: \(q_{\text{heating steam}} = 18.0 \ \text{g} \times 2.02 \ \frac{\text{J}}{^\circ \text{C} \cdot \text{g}} \times 40 \ ^\circ\text{C} = 1454.4 \ \text{J}\) Total heat: \(q_{\text{total}} = 1096.2 \ \text{J} + 6020 \ \text{J} + 7532.4 \ \text{J} + 40700 \ \text{J} + 1454.4 \ \text{J} = 56803 \ \text{J}\)

Step by step solution

01

Heating the ice

First, we need to heat the ice from its initial temperature of -30°C to its melting point of 0°C. To find the amount of heat required for this, we can use the specific heat capacity of ice (c_ice) and the following formula: \[q_{\text{heating ice}} = m \times c_\text{ice}\times \Delta T\] where m is the mass (in g) of the ice and ΔT is the change in temperature. Since we have 1 mole of water, the water mass is 18.0 grams (1.00 mole × 18.0 g/mol). We have, c_ice = 2.03 J/g°C ΔT = 0 - (-30) = 30°C So, \[q_{\text{heating ice}} = 18.0 \ \text{g} \times 2.03 \ \frac{\text{J}}{^\circ \text{C} \cdot \text{g}} \times 30 \ ^\circ\text{C}\]
02

Melting the ice

Next, we determine the amount of heat needed to melt the ice at 0°C. To find this, we multiply the enthalpy of fusion (ΔH_fus) with the moles of water: \[q_{\text{melting ice}} = 1.00 \ \text{mol} \times 6.02 \ \frac{\text{kJ}}{\text{mol}} \times \frac{1000 \ \text{J}}{\text{kJ}}\]
03

Heating the water

Now, we calculate the heat needed to heat the liquid water from 0°C to its boiling point at 100°C. To do this, we use the specific heat capacity of water (c_water) and the same formula as in step 1: \[q_{\text{heating water}} = m \times c_\text{water} \times \Delta T\] We have, c_water = 4.18 J/g°C ΔT = 100 - 0 = 100°C So, \[q_{\text{heating water}} = 18.0 \ \text{g} \times 4.18 \ \frac{\text{J}}{^\circ \text{C} \cdot \text{g}} \times 100 \ ^\circ\text{C}\]
04

Vaporizing the water

Next, we determine the heat needed to vaporize the water at 100°C. To find this, we multiply the enthalpies of vaporization (ΔH_vap) with the moles of water: \[q_{\text{vaporizing water}} = 1.00 \ \text{mol} \times 40.7 \ \frac{\text{kJ}}{\text{mol}} \times \frac{1000 \ \text{J}}{\text{kJ}}\]
05

Heating the steam

Finally, we need to heat the steam from 100°C to its final temperature of 140°C. To do this, we use the specific heat capacity of steam (c_steam) and the same formula as in step 1: \[q_{\text{heating steam}} = m \times c_\text{steam} \times \Delta T\] We have, c_steam = 2.02 J/g°C ΔT = 140 - 100 = 40°C So, \[q_{\text{heating steam}} = 18.0 \ \text{g} \times 2.02 \ \frac{\text{J}}{^\circ \text{C} \cdot \text{g}} \times 40 \ ^\circ\text{C}\]
06

Total heat

Finally, we sum up the heats from all stages: \[q_{\text{total}} = q_{\text{heating ice}} + q_{\text{melting ice}} + q_{\text{heating water}} + q_{\text{vaporizing water}} + q_{\text{heating steam}}\] Calculate each term from Steps 1-5 and sum them up to find the total heat q.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Heat Capacity
Understanding specific heat capacity is crucial when we want to determine how much energy it takes to change the temperature of a substance. It is defined as the amount of heat per unit mass needed to raise the temperature by 1 degree Celsius. In simpler terms, it measures a substance's 'heat tolerance'. The higher the specific heat capacity, the more energy it can absorb without significantly changing its temperature.

To calculate the energy involved in changing the temperature of a substance, we use the formula:
\[q = m \times c \times \Delta T\]
Where \(q\) is the heat absorbed or released (in joules, J), \(m\) is the mass of the substance (in grams, g), \(c\) is the specific heat capacity (in J/g°C), and \(\Delta T\) is the change in temperature (in °C). This formula is utilized at different stages of the thermochemical process to ascertain the heat required to raise the temperature of various forms of water (ice, liquid, and steam) in our original exercise.
Enthalpy of Fusion
The enthalpy of fusion, also known as the heat of fusion, is the energy required to change a substance from solid to liquid at its melting point, without any temperature change. This is a type of latent heat, which is related to the phase changes of a substance rather than its temperature change. For water, this occurs at 0°C, and the energy required to melt ice is considerable.

The formula to calculate the heat needed for the phase change from solid to liquid is:
\[q_{\text{fusion}} = n \times \Delta H_{\text{fusion}}\]
Here, \(n\) represents the number of moles, and \(\Delta H_{\text{fusion}}\) is the enthalpy of fusion per mole. When we perform this calculation for our exercise example, we're able to find out how much energy it takes to melt a mole of ice at its melting point.
Enthalpy of Vaporization
Similarly to the enthalpy of fusion, the enthalpy of vaporization signifies the energy needed to change a substance from liquid to gas at its boiling point. It once again reflects the concept of latent heat but is usually much greater than the heat of fusion because breaking intermolecular forces in a liquid to make it into a gas requires more energy.

The formula to calculate this phase change is:
\[q_{\text{vaporization}} = n \times \Delta H_{\text{vaporization}}\]
For our exercise example, this calculation shows the significant amount of heat necessary to convert liquid water to steam at 100°C. This step is one of the key stages when heating water from a solid state below freezing to a gaseous state above boiling.
Temperature Conversion
Temperature conversion is important when we're dealing with temperature scales different from the one we're accustomed to. However, in thermochemistry, temperatures are often presented in degrees Celsius or Kelvin, and it is not typically necessary to convert between these two scales, as a change in temperature by 1 degree Celsius is equivalent to a change of 1 Kelvin.

Nonetheless, it might be helpful to know how to convert degrees Celsius to Fahrenheit in some contexts, using the formula:\[F = \frac{9}{5}C + 32\]
where \(F\) represents degrees Fahrenheit and \(C\) represents degrees Celsius. This is not directly relevant to our original exercise, which remains in the Celsius scale, but is useful information for comprehensive understanding.
Heat Transfer
Heat transfer pertains to the movement of thermal energy from one object or material to another. Three modes of heat transfer exist: conduction, convection, and radiation. When we heat water from solid ice at -30°C to gaseous steam at 140°C, we apply energy that moves from the heat source into the water through conduction. The water molecules absorb the energy, causing temperature changes and phase transitions, illustrating heat transfer in action.

In our original exercise, understanding the heat transfer concept is fundamental as it allows comprehension of each step involved in the heating process, and why sufficient energy (heat) must be provided to overcome the thermal inertia (specific heat) and phase-change resistances (enthalpies of fusion and vaporization) that water possesses.

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Most popular questions from this chapter

The enthalpy of combustion of \(\mathrm{CH}_{4}(g)\) when \(\mathrm{H}_{2} \mathrm{O}(l)\) is formed is \(-891 \mathrm{~kJ} / \mathrm{mol}\) and the enthalpy of combustion of \(\mathrm{CH}_{4}(\mathrm{~g})\) when \(\mathrm{H}_{2} \mathrm{O}(g)\) is formed is \(-803 \mathrm{~kJ} / \mathrm{mol}\). Use these data and Hess's law to determine the enthalpy of vaporization for water.

Consider a mixture of air and gasoline vapor in a cylinder with a piston. The original volume is \(40 . \mathrm{cm}^{3}\). If the combustion of this mixture releases 950. J of energy, to what volume will the gases expand against a constant pressure of 650 . torr if all the energy of combustion is converted into work to push back the piston?

Photosynthetic plants use the following reaction to produce glucose, cellulose, and so forth: $$ 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \stackrel{\text { Sunlight }}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) $$ How might extensive destruction of forests exacerbate the greenhouse effect?

The preparation of \(\mathrm{NO}_{2}(g)\) from \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) is an endothermic reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) \text { (unbalanced) } $$ The enthalpy change of reaction for the balanced equation (with lowest whole- number coefficients) is \(\Delta H=67.7 \mathrm{~kJ}\). If \(2.50 \times 10^{2} \mathrm{~mL} \mathrm{~N}_{2}(g)\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(3.50 \mathrm{~atm}\) and \(4.50 \times\) \(10^{2} \mathrm{~mL} \mathrm{O}_{2}(g)\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(3.50\) atm are mixed, what amount of heat is necessary to synthesize the maximum yield of \(\mathrm{NO}_{2}(g) ?\)

A gas absorbs \(45 \mathrm{~kJ}\) of heat and does \(29 \mathrm{~kJ}\) of work. Calculate \(\Delta E\).
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