Problem 48 - University Physics with Modern Physics

Question

An ice-cule tray of negligible mass contains 0.350 \(\mathrm{kg}\) of water at \(18.0^{\circ} \mathrm{C} .\) How much heat must be removed to cool the water to \(0.00^{\circ} \mathrm{C}\) and freeze it? Express your answer in joules, calories, and Btu.

Step-by-Step Solution

Answer

The total heat removed is approximately 143343 J, 34253 cal, and 136 BTU.

Step 1: Determine the Heat to Cool Water to Freezing Point

First, calculate the amount of heat required to cool the water from its initial temperature to its freezing point, which is 0°C. The specific heat capacity of water is. The formula to calculate heat is: \[ Q_1 = m imes c imes riangle T \]where \( m = 0.350 \text{ kg} \), \( c = 4.186 \text{ J/g°C} \), and \( \triangle T = 18.0°C - 0°C \). Note that 1 kg = 1000 g. So, substitute these values in:\[ Q_1 = 350 \times 4.186 \times 18.0 = 26442.6 \text{ J} \]

Step 2: Determine the Heat to Freeze the Water

Now calculate the heat required to freeze the water. The latent heat of fusion for water is 334 J/g. Use the formula:\[ Q_2 = m \times L_f \]where \( L_f = 334 \text{ J/g} \). Substitute the values into the equation:\[ Q_2 = 350 \times 334 = 116900 \text{ J} \]

Step 3: Calculate the Total Heat Removed

Add the heat needed to cool the water to 0°C (\( Q_1 \)) and the heat required to freeze the water (\( Q_2 \)) to find the total heat removed:\[ Q_{total} = Q_1 + Q_2 = 26442.6 + 116900 = 143342.6 \text{ J} \]

Step 4: Convert Joules to Calories

To convert the total heat from joules to calories, use the conversion factor: 1 calorie = 4.184 joules. Therefore:\[ Q_{total} = \frac{143342.6}{4.184} \approx 34252.7 \text{ cal} \]

Step 5: Convert Joules to BTU

To convert joules to British Thermal Units (BTU), use the conversion factor: 1 BTU = 1055.06 joules. Thus:\[ Q_{total} = \frac{143342.6}{1055.06} \approx 135.9 \text{ BTU} \]

Key Concepts

Understanding Specific Heat Capacity

Specific heat capacity is a fundamental concept in thermodynamics that represents the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. For water, this value is quite significant at 4.186 J/g°C. This means that water can absorb a lot of heat before its temperature changes.
For practical purposes, the specific heat capacity tells us how much energy we need to add or remove to change the temperature of a given mass of a substance. In the exercise we dealt with, the specific heat capacity of water helps calculate how much heat is needed to lower the water's temperature from 18.0°C to 0.0°C.

• The formula used is: \[ Q = m \times c \times \Delta T \] Here, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.

Through this relationship, the energy required to bring the water to its freezing point was found to be 26442.6 J, showcasing the high capacity for water to store thermal energy.

The Concept of Latent Heat of Fusion

Latent heat of fusion is another critical thermodynamic concept that explains the energy required to change a substance from solid to liquid or vice versa at constant temperature. For water, the latent heat of fusion is 334 J/g, which means that 334 joules of energy are needed to melt or freeze one gram of ice without changing its temperature.
In our exercise, after cooling the water to 0°C, the next step is freezing it. The heat removed during this phase change does not decrease the temperature further but only changes the state from liquid to solid.

• The formula to determine the heat for the phase change: \[ Q = m \times L_f \] \(L_f\) is the latent heat of fusion.

To complete the phase change for the example problem, 116900 J of heat was removed, allowing the water to transition into ice perfectly.

Exploring Heat Energy Conversion

Heat energy conversion is essential for understanding various energy units and making useful comparisons between them. In our exercise, once the total heat was found, converting it into different units allowed for universal understanding and comparison.
Firstly, converting joules to calories, where the relationship is 1 calorie = 4.184 joules, allows us to express heat in a more human-scale unit. This conversion showed that we needed about 34252.7 calories for our task.
Additionally, converting to British Thermal Units (BTU) is useful, especially in regions and industries where BTU is a standard unit for measuring heat. This involves knowing that 1 BTU = 1055.06 joules, translating our example's required energy into approximately 135.9 BTU.

• Formulas for conversion include:
  \[ \text{calories} = \frac{\text{joules}}{4.184} \]\ \[ \text{BTU} = \frac{\text{joules}}{1055.06} \]

These conversions allow for diverse applications and better communication of thermal energy across different scientific and engineering fields.