Problem 51 - Chemistry The Central Science

Question

(a) What is the specific heat of liquid water? (b) What is the molar heat capacity of liquid water? (c) What is the heat capacity of \(185 \mathrm{~g}\) of liquid water? (d) How many \(\mathrm{kJ}\) of heat are needed to raise the temperature of \(10.00 \mathrm{~kg}\) of liquid water from \(24.6^{\circ} \mathrm{C}\) to \(46.2^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

(a) The specific heat of liquid water is \(4.18 \,\text{J/g}^\circ\text{C}\). (b) The molar heat capacity of liquid water is approximately \(75.33 \,\text{J/mol}^\circ\text{C}\). (c) The heat capacity of 185 g of liquid water is approximately \(773.3 \,\text{J/}^\circ\text{C}\). (d) The heat needed to raise the temperature of 10 kg of liquid water from 24.6°C to 46.2°C is approximately \(903.36 \,\text{kJ}\).

Step 1: (a) Specific heat of liquid water

The specific heat of liquid water is \(4.18 \,\text{J/g}^\circ\text{C}\). This value is experimentally determined and can be found in reference materials such as textbooks and online resources.

Step 2: (b) Molar heat capacity of liquid water

To find the molar heat capacity of liquid water, we need to multiply the specific heat by the molar mass of water. The molar mass of water is approximately 18.015 g/mol. Molar heat capacity = Specific heat × Molar mass Molar heat capacity = \(4.18 \,\text{J/g}^\circ\text{C} \times 18.015\,\text{g/mol}\) Molar heat capacity ≈ \(75.33 \,\text{J/mol}^\circ\text{C}\)

Step 3: (c) Heat capacity of 185 g of liquid water

To calculate the heat capacity of 185 g of liquid water, we use the formula: Heat capacity = Mass × Specific heat Heat capacity = \(185\,\text{g} \times 4.18 \,\text{J/g}^\circ\text{C}\) Heat capacity ≈ \(773.3 \,\text{J/}^\circ\text{C}\)

Step 4: (d) Heat needed to raise the temperature of 10 kg of liquid water

To calculate the heat needed to raise the temperature of 10 kg of liquid water from 24.6°C to 46.2°C, we use the heat transfer formula: Q = mcΔT where Q is the heat needed, m is the mass of water, c is the specific heat, and ΔT is the change in temperature. First, we need to convert the mass of water from kg to g: Mass of water = \(10,000\,\text{g}\) (since 1 kg = 1000 g) Next, we calculate the change in temperature: ΔT = T_final - T_initial ΔT = 46.2°C - 24.6°C ΔT = 21.6°C Now, we can plug in the values into the heat transfer formula: Q = \(10,000\,\text{g} \times 4.18 \,\text{J/g}^\circ\text{C} \times 21.6^\circ\text{C}\) Q ≈ \(903,360 \,\text{J}\) To convert the heat needed to kilojoules, we divide by 1000: Q ≈ \(903.36 \,\text{kJ}\)

Key Concepts

Molar Heat Capacity

Molar heat capacity is an interesting concept that represents the amount of heat needed to change the temperature of one mole of a substance by one degree Celsius. It is directly related to specific heat, which is the heat required for a single gram of a substance. For water, this can be determined using its specific heat and molar mass.
To calculate the molar heat capacity (\( C_m \)) of water, you multiply its specific heat (\( c_p \)) by its molar mass (\( M \)). The expression becomes:

• \( C_m = c_p \times M \)

Water's specific heat is 4.18 J/g°C and its molar mass is about 18.015 g/mol. By multiplying these two,
we find:

• \( C_m = 4.18 \, ext{J/g}^\circ\text{C} \times 18.015 \,\text{g/mol} \approx 75.33 \,\text{J/mol}^\circ\text{C} \)

This value reflects how much heat energy is needed to raise 1 mole of liquid water by 1°C, providing a more scalable insight compared to specific heat.

Heat Transfer

Heat transfer in the context of temperature change is a critical process in thermodynamics, describing how heat energy moves from one place to another. The formula for calculating the heat (\( Q \)) required for a temperature change involves a few variables: the mass of the substance (\( m \)), its specific heat (\( c \)), and the change in temperature (\( \Delta T \)). The equation is expressed as:

• \( Q = mc\Delta T \)

The specific heat acts like a coefficient to resist temperature change, while the mass and temperature difference define the overall capacity for heat absorption or release.
In the context of water,
if you need to increase the temperature of 10 kg of water from 24.6°C to 46.2°C:

• Mass (\( m \)) = 10,000 g
• Specific heat (\( c \)) = 4.18 J/g°C
• Temperature change (\( \Delta T \)) = 21.6°C

Inserting these in the formula gives:

• \( Q = 10,000 \,\text{g} \times 4.18 \,\text{J/g}^\circ\text{C} \times 21.6^\circ\text{C} \approx 903,360 \,\text{J} \)

This considerable amount of energy demonstrates how liquid water, with its high specific heat, demands substantial energy even for modest temperature changes.

Temperature Change

Temperature change (\( \Delta T \)) is an essential variable in heat transfer calculations, indicating how much a substance's temperature is altered when heat is applied. The precision and magnitude of this change depend on the material's specific heat capacity.
Water, with its high specific heat, absorbs heat efficiently without a large temperature increase. When computing a temperature change, we identify both the initial and final temperatures and subtract them:

• \( \Delta T = T_{\text{final}} - T_{\text{initial}} \)

For example, if you have water initially at 24.6°C and it heats up to 46.2°C, the calculated change would be:

• \( \Delta T = 46.2^\circ\text{C} - 24.6^\circ\text{C} = 21.6^\circ\text{C} \)

This value tells us the level of thermal energy discrepancy facilitated by the applied heat, aiding in understanding the overall balance of heat in reactions or processes. It highlights the resilience of water to temperature shifts due to its intrinsic properties.