/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 The following table gives the am... [FREE SOLUTION] | 91Ó°ÊÓ

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The following table gives the amount of energy required in order to heat, melt, or boil a gram of water. $$\begin{array}{ll}\hline \text { heat } 1 \mathrm{~g} \text { of ice by } 1^{\circ} \mathrm{C} & 2.05 \mathrm{~J} \\ \text { melt } 1 \mathrm{~g} \text { of ice } & 333 \mathrm{~J} \\ \text { heat } 1 \mathrm{~g} \text { of liquid by } 1^{\circ} \mathrm{C} & 4.19 \mathrm{~J} \\ \text { boil } 1 \mathrm{~g} \text { of water } & 2500 \mathrm{~J} \\ \text { heat } 1 \mathrm{~g} \text { of steam by } 1^{\circ} \mathrm{C} & 2.01 \mathrm{~J} \\ \hline\end{array}$$ (a) How much energy is required in order to convert \(1.00 \mathrm{~g}\) of ice at \(-20^{\circ} \mathrm{C}\) into steam at \(137^{\circ} \mathrm{C}\) ? (answer check available at lightandmatter.com) (b) What is the minimum amount of hot water that could melt \(1.00 \mathrm{~g}\) of ice? (answer check available at lightandmatter.com)

Short Answer

Expert verified
(a) 3367 J to convert 1g of ice at -20°C to steam at 137°C. (b) Minimum 0.80 g of hot water needed to melt 1g ice.

Step by step solution

01

Heating ice from -20°C to 0°C

The specific heat capacity of ice is 2.05 J/g°C. We need to heat 1 g of ice from -20°C to 0°C. The change in temperature is 20°C.The energy required is calculated by the formula: \( q = m \cdot c \cdot \Delta T \)\( q = 1 \text{g} \times 2.05 \text{J/g°C} \times 20°C = 41 \text{J} \).
02

Melting ice at 0°C

To melt 1 g of ice at 0°C, we use the heat of fusion for ice, which is 333 J/g.The energy required is:\( q = 333 \text{J} \).
03

Heating water from 0°C to 100°C

The specific heat capacity of liquid water is 4.19 J/g°C. We need to heat 1 g of water from 0°C to 100°C. The change in temperature is 100°C.The energy required is calculated by:\( q = m \cdot c \cdot \Delta T \)\( q = 1 \text{g} \times 4.19 \text{J/g°C} \times 100°C = 419 \text{J} \).
04

Boiling water at 100°C

To convert 1 g of water to steam at 100°C, we use the heat of vaporization, which is 2500 J/g.The energy required is:\( q = 2500 \text{J} \).
05

Heating steam from 100°C to 137°C

The specific heat capacity of steam is 2.01 J/g°C. We need to heat 1 g of steam from 100°C to 137°C. The change in temperature is 37°C.The energy required is calculated by:\( q = m \cdot c \cdot \Delta T \)\( q = 1 \text{g} \times 2.01 \text{J/g°C} \times 37°C = 74.37 \text{J} \).
06

Total Energy Calculation for Part (a)

Add up all the energies from Steps 1-5:\( 41 \text{J} + 333 \text{J} + 419 \text{J} + 2500 \text{J} + 74.37 \text{J} = 3367.37 \text{J} \).Therefore, the total energy required is approximately 3367 J.
07

Calculating Energy Required to Melt 1g of Ice for Part (b)

To melt 1 g of ice, 333 J of energy is required (as found in Step 2).
08

Calculating Minimum Hot Water Needed for Part (b)

Let the mass of hot water be \( m \). Assume hot water is initially at 100°C and is cooled to 0°C. Using specific heat of water (4.19 J/g°C), the energy released is:\( 4.19 \text{J/g°C} \times 100°C \times m \text{g} \).Set this equal to 333 J:\( 4.19 \times 100 \times m = 333 \ m = \frac{333}{419} \approx 0.795 \text{ g} \).Therefore, a minimum of approximately 0.80 g of hot water is needed.

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is a measure of the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius. This property is crucial in understanding how different materials absorb and retain heat. For instance, water is known for its high specific heat capacity of 4.19 J/g°C, which means it requires a lot of energy to change temperature. Conversely, ice has a specific heat capacity of 2.05 J/g°C, indicating it requires less energy compared to liquid water. Understanding this concept helps in calculating how much energy is needed to heat or cool a substance under different conditions.
Heat of Fusion
The heat of fusion refers to the energy required to change a substance from solid to liquid at its melting point, without changing its temperature. In the case of ice, the heat of fusion is significant, as it is 333 J/g. This means that to melt one gram of ice at 0°C into water, you need to provide 333 Joules of energy. This concept is particularly important in processes where ice transitions to water, such as in refrigeration or even natural processes like the melting of ice caps. Understanding the heat of fusion allows us to calculate how much energy is involved in melting and freezing processes.
Heat of Vaporization
The heat of vaporization is the amount of energy needed to turn a liquid into a gas at its boiling point, without changing its temperature. For water, this is a whopping 2500 J/g. This means transforming one gram of liquid water at 100°C into steam requires 2500 Joules of energy. This large energy requirement is due to the strong hydrogen bonds that need to be broken for water molecules to move from the liquid to the gaseous state. This concept is crucial in many industrial processes, such as distillation and power generation, where heat of vaporization plays a fundamental role in energy and material transformation.
Phase Changes
Phase changes involve the transformation of a substance from one state of matter to another, such as from solid to liquid or liquid to gas. Each phase change occurs at specific temperatures and involves absorbing or releasing energy while the temperature remains constant. Water's transitions, from ice (solid) to liquid water and then to steam (gas), illustrate the nature of phase changes. During these transitions, the temperature doesn't increase even though energy is absorbed—showcasing the distinction between temperature change and phase change. Understanding phase changes is essential in fields like climatology and culinary science, where precise control of temperature and state is necessary.
Energy Calculations
Energy calculations in thermodynamics involve determining how much energy is required or released during heating, cooling, and phase changes. For example, calculating the total energy needed to convert ice at -20°C to steam at 137°C involves several steps. First, calculate the energy needed to heat the ice up to 0°C, then the energy to melt the ice, followed by the energy required to heat the resulting water to 100°C. After boiling, calculate the energy required to heat the obtained steam to the final temperature. Each step involves a specific formula tying mass, specific heat capacity, and temperature change together, often summarized as: \[ q = m \cdot c \cdot \Delta T \] for simple heating or cooling, and using specific enthalpy values for phase changes. By adding up all energies involved, you derive the total energy required for complex thermodynamic processes. This skill enables you to tackle problems in engineering, environmental science, and everyday applications like cooking.

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Most popular questions from this chapter

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