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When \(5.0 \mathrm{~g}\) of a certain type of coal is burned, it raises the temperature of \(1000 \mathrm{~mL}\) of water from \(10{ }^{\circ} \mathrm{C}\) to \(47^{\circ} \mathrm{C}\). Calculate the thermal energy produced per gram of coal. Neglect the small heat capacity of the coal.

Short Answer

Expert verified
The thermal energy produced per gram of coal is approximately 30,932 J/g.

Step by step solution

01

Understand the Problem

The problem asks us to calculate the thermal energy produced per gram of coal when a certain mass of coal is burned and raises the temperature of a known volume of water.
02

Identify Known Variables

We are given the following information:- Mass of water, \( m = 1000 \text{ mL} = 1000 \text{ g} \) (since the density of water \( \approx 1 \text{ g/mL} \))- Initial temperature of water, \( T_i = 10^{\circ} \text{C} \)- Final temperature of water, \( T_f = 47^{\circ} \text{C} \)- Mass of coal burned, \( m_{\text{coal}} = 5.0 \text{ g} \)
03

Determine the Temperature Change

Calculate the temperature change of the water:\[ \Delta T = T_f - T_i = 47^{\circ} \text{C} - 10^{\circ} \text{C} = 37^{\circ} \text{C} \]
04

Use the Specific Heat Formula

The formula to calculate heat absorbed by the water is given by:\[ q = m \cdot c \cdot \Delta T \]where:- \( m \) is the mass of the water- \( c \) is the specific heat capacity of water (\(4.18 \text{ J/g} \cdot ^{\circ}C \))- \( \Delta T \) is the temperature change
05

Calculate the Heat Energy Absorbed by the Water

Substituting the known values into the formula:\[ q = 1000 \text{ g} \cdot 4.18 \text{ J/g} \cdot ^{\circ}C \cdot 37^{\circ}C = 154,660 \text{ J} \]
06

Calculate the Thermal Energy Produced Per Gram of Coal

Divide the total energy by the mass of coal to find energy per gram:\[ \text{Energy per gram of coal} = \frac{154,660 \text{ J}}{5.0 \text{ g}} = 30,932 \text{ J/g} \]
07

Final Calculation and Answer

We have calculated the energy produced as \(30,932 \text{ J/g}\) using the formula for heat and dividing by the mass of coal burned.

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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 vital element in understanding how different substances absorb heat. It is the amount of energy required to raise the temperature of one gram of a substance by one degree Celsius.
For example, water has a specific heat capacity of \( 4.18 \text{ J/g} \cdot ^{\circ}C \). This means it takes 4.18 joules of energy to increase the temperature of 1 gram of water by 1 degree Celsius. This high value indicates that water can store a lot of heat energy.
Knowing the specific heat capacity is essential for calculating how much energy is absorbed by or released from a substance. It helps us understand energy changes in chemical reactions or physical processes involving heat transfer.
Heat Transfer
Heat transfer is the movement of thermal energy from one object to another. It occurs when there is a temperature difference. The heat will naturally flow from the hotter object to the cooler one until equilibrium is reached.
In our exercise, the heat transfer is from the burning coal to the water, raising the temperature of the water. It's calculated using formulas like \( q = m \cdot c \cdot \Delta T \), where \( q \) is the amount of heat transferred, \( m \) is the mass, \( c \) is specific heat capacity, and \( \Delta T \) is the temperature change.
  • Conduction: Transfer through direct contact, like a metal spoon in hot coffee.
  • Convection: Transfer through fluid movements, like boiling water.
  • Radiation: Transfer through electromagnetic waves, like the sun warming your face.
Understanding heat transfer helps us design better systems for heating and cooling, improving energy efficiency.
Temperature Change
Temperature change refers to the difference in temperature between the final and initial states of a system. It is often denoted as \( \Delta T \), and in our example, the water's temperature increases by \( 37^{\circ}C \).
This concept is essential for calculating the energy exchange in systems. When heat is added or removed, the temperature of a substance changes according to its specific heat capacity and mass.
In mathematical terms, you compute the change with \( \Delta T = T_f - T_i \), where \( T_f \) is the final temperature, and \( T_i \) is the initial temperature. This calculation helps in determining whether a chemical reaction or a physical process has absorbed or released energy.
Mass-Energy Relationship
The mass-energy relationship is a fundamental concept that shows how mass affects the energy required for temperature changes. In our problem, 5 grams of coal is burned to raise the temperature of 1000 grams of water.
The energy released depends on the mass of the coal and water. More mass typically requires more energy to change its temperature, further illustrated by the formula \( q = m \cdot c \cdot \Delta T \).
This relationship is crucial in fields like chemistry and physics, helping explain phenomena such as combustion and fusion. It emphasizes how mass directly influences energy required or produced during reactions and processes, providing deeper insights into energy conservation and utilization.

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Most popular questions from this chapter

In a calorimeter can (which behaves thermally as if it were equivalent to \(40 \mathrm{~g}\) of water) are \(200 \mathrm{~g}\) of water and \(50 \mathrm{~g}\) of ice, all at exactly \(0{ }^{\circ} \mathrm{C}\). Into this is poured \(30 \mathrm{~g}\) of water at \(90{ }^{\circ} \mathrm{C}\). What will be the final condition of the system? Let us start by assuming (perhaps incorrectly) that the final temperature is \(T_{f}>0{ }^{\circ} \mathrm{C}\). Then $$ \begin{array}{c} \left(\begin{array}{l} \text { Heat change of } \\ \text { hot water } \end{array}\right)+\left(\begin{array}{l} \text { Heat to } \\ \text { melt ice } \end{array}\right)+\left(\begin{array}{l} \text { Heat to warm } \\ 250 \mathrm{~g} \text { of water } \end{array}\right)+\left(\begin{array}{l} \text { Heat to warm } \\ \text { calorimeter } \end{array}\right)=0 \\ (30 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(T_{f}-90{ }^{\circ} \mathrm{C}\right)+(50 \mathrm{~g})(80 \mathrm{cal} / \mathrm{g})+(250 \mathrm{~g})\left(1 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(T_{f}-0{ }^{\circ} \mathrm{C}\right) \\ +(40 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(T_{f}-0{ }^{\circ} \mathrm{C}\right)=0 \end{array} $$ Solving gives \(T_{f}=-4.1^{\circ} \mathrm{C}\), contrary to our assumption that the final temperature is above \(0{ }^{\circ} \mathrm{C}\). Apparently, not all the ice melts. Therefore, \(T_{f}=0{ }^{\circ} \mathrm{C}\). To find how much ice melts, we write Heat lost by hot water \(=\) Heat gained by melting ice $$ (30 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(90{ }^{\circ} \mathrm{C}\right)=(80 \mathrm{cal} / \mathrm{g}) m $$ where \(m\) is the mass of ice that melts. Solving this equation yields \(m=34 \mathrm{~g}\). The final system has \(50 \mathrm{~g}-34 \mathrm{~g}=16 \mathrm{~g}\) of ice not melted.

Outside air at \(5{ }^{\circ} \mathrm{C}\) and 20 percent relative humidity is introduced into a heating and air-conditioning plant where it is heated to \(20^{\circ} \mathrm{C}\) and its relative humidity is increased to a comfortable 50 percent. How many grams of water must be evaporated into a cubic meter of outside air to accomplish this? Saturated air at \(5^{\circ} \mathrm{C}\) contains \(6.8 \mathrm{~g} / \mathrm{m}^{3}\) of water, and at \(20^{\circ} \mathrm{C}\) it contains \(17.3 \mathrm{~g} / \mathrm{m}^{3}\). $$ \begin{array}{l} \text { Mass } / \mathrm{m}^{3} \text { of water vapor in air at } 5^{\circ} \mathrm{C}=0.20 \times 6.8 \mathrm{~g} / \mathrm{m}^{3}=1.36 \mathrm{~g} / \mathrm{m}^{3}\\\ \text { Comfortable mass } / \mathrm{m}^{3} \text { at } 20^{\circ} \mathrm{C}=0.50 \times 17.3 \mathrm{~g} / \mathrm{m}^{3}=8.65 \mathrm{~g} / \mathrm{m}^{3}\\\ 1 \mathrm{~m}^{3} \text { of air at } 5^{\circ} \mathrm{C} \text { expands to }(293 / 278) \mathrm{m}^{3}=1.054 \mathrm{~m}^{3} \text { at } 20^{\circ} \mathrm{C}\\\ \text { Mass of water vapor in } 1.054 \mathrm{~m}^{3} \text { at } 20^{\circ} \mathrm{C}=1.054 \mathrm{~m}^{3} \times 8.65 \mathrm{~g} / \mathrm{m}^{3}=9.12 \mathrm{~g}\\\ \text { Mass of water to be added to each } m^{3} \text { of air at } 5^{\circ} \mathrm{C}=(9.12-1.36) \mathrm{g}=7.8 \mathrm{~g} \end{array} $$

Determine the temperature \(T_{f}\) that results when \(150 \mathrm{~g}\) of ice at \(0{ }^{\circ} \mathrm{C}\) is mixed with \(300 \mathrm{~g}\) of water at \(50^{\circ} \mathrm{C}\). From energy conservation, $$ \begin{array}{r} \text { (Heat change of ice) }+\text { (Heat change of water) }=0 \\ \text { (Heat to melt ice) }+\text { (Heat to warm ice water) }+(\text { Heat change of water })=0 \\ \left(m L_{f \text { ice }}+\left(\text { (cm } \Delta T_{\text {ice water }}+(c m \Delta T)_{\text {water }}=0\right.\right. \\ (150 \mathrm{~g})(80 \mathrm{cal} / \mathrm{g})+\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(150 \mathrm{~g})\left(T_{f}-0^{\circ} \mathrm{C}\right)+\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(300 \mathrm{~g})\left(T_{f}-50^{\circ} \mathrm{C}\right)=0 \end{array} $$ from which \(T_{f}=6.7^{\circ} \mathrm{C}\)

Suppose a person who eats 2500 Cal of food each day loses the heat equivalent of the food through evaporation of water from the body. How much water must evaporate each day? At body temperature, \(L_{v}\) for water is about \(600 \mathrm{cal} / \mathrm{g}\)

On a certain day the temperature is \(20^{\circ} \mathrm{C}\) and the dew point is \(5.0{ }^{\circ} \mathrm{C}\). What is the relative humidity? Saturated air at \(20^{\circ} \mathrm{C}\) and \(5.0{ }^{\circ} \mathrm{C}\) contains \(17.12\) and \(6.80 \mathrm{~g} / \mathrm{m}^{3}\) of water, respectively.

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