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What quantity of energy, in joules, is required to raise the temperature of \(454 \mathrm{g}\) of tin from room temperature, \(25.0^{\circ} \mathrm{C},\) to its melting point, \(231.9^{\circ} \mathrm{C},\) and then melt the tin at that temperature? (The specific heat capacity of tin is \(0.227 \mathrm{J} / \mathrm{g} \cdot \mathrm{K},\) and the heat of fusion of this metal is \(59.2 \mathrm{J} / \mathrm{g} .\) )

Short Answer

Expert verified
The total energy required is approximately 48224 J.

Step by step solution

01

Identify Given Data

We are given the following information: the mass of tin is \(454 \, \mathrm{g}\), the initial temperature is \(25.0^{\circ} \mathrm{C}\), the melting point is \(231.9^{\circ} \mathrm{C}\), the specific heat capacity of tin \(c\) is \(0.227 \, \mathrm{J/g \cdot K}\), and the heat of fusion \( \Delta H_f\) is \(59.2 \, \mathrm{J/g}\).
02

Calculate Temperature Change

The change in temperature (\(\Delta T\)) is the final temperature minus the initial temperature: \(\Delta T = 231.9^{\circ} \mathrm{C} - 25.0^{\circ} \mathrm{C} = 206.9 \, \mathrm{K}\).
03

Calculate Energy to Heat Tin

The amount of energy needed to heat the tin to its melting point is given by \(Q = mc\Delta T\). Substituting the known values, \(Q = 454 \, \mathrm{g} \times 0.227 \, \mathrm{J/g \cdot K} \times 206.9 \, \mathrm{K} = 21347.67 \, \mathrm{J}\).
04

Calculate Energy to Melt Tin

The energy required to melt the tin is calculated using the heat of fusion \(Q = m \Delta H_f\). So, \(Q = 454 \, \mathrm{g} \times 59.2 \, \mathrm{J/g} = 26876.8 \, \mathrm{J}\).
05

Calculate Total Energy Required

The total energy required is the sum of the energy needed to raise the temperature and the energy to melt the tin. Therefore, \(21427.67 \, \mathrm{J} + 26876.8 \, \mathrm{J} = 48224.47 \, \mathrm{J}\).
06

Rounding the Final Result

Round the total energy to the appropriate number of significant figures. The final total energy required is approximately \(48224.47 \, \mathrm{J}\).

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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 crucial concept in understanding how heat energy is absorbed by substances. It measures the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin). In this context, the specific heat capacity of tin is given as \(0.227 \ \mathrm{J/g \cdot K}\).
This value tells us how much energy, in joules, is needed per gram of tin to achieve a temperature increase of one Kelvin. Knowing the specific heat capacity helps predict how a material reacts to changes in heat.
  • A higher specific heat capacity means the substance will absorb more heat without a significant temperature change.
  • A lower specific heat capacity indicates that less heat is required for the same temperature increase, making the substance heat up quickly.
For our exercise, we used this in the formula \(Q = mc\Delta T\) to compute the energy required to increase the temperature of tin from room temperature to its melting point.
heat of fusion
The heat of fusion is an essential thermal property that indicates the amount of energy needed to change a substance from solid to liquid at its melting point, without altering its temperature.
For tin, the heat of fusion is provided as \(59.2 \ \mathrm{J/g}\). This means that each gram of tin requires \(59.2 \ \mathrm{J}\) of energy to transform from a solid to a liquid state at 231.9°C.
  • The heat of fusion is particularly important for substances that transition between solid and liquid phases, such as metals like tin.
  • Understanding this concept ensures that the necessary energy calculations account for phase changes, not just temperature increases.
In the exercise, the formula \(Q = m \Delta H_f\) was used to determine the energy needed to melt the tin, showing how the heat of fusion directly relates to energy changes.
temperature change
Temperature change is a fundamental part of calculating energy requirements in thermodynamics. It's the difference between the initial and final temperatures of a substance. Here, the change in temperature for tin was from \(25.0^{\circ} \mathrm{C}\) to \(231.9^{\circ} \mathrm{C}\), which is a total change of \(206.9 \ \mathrm{K}\).
Temperature change is crucial because it impacts the amount of heat involved in heating the substance.
  • A larger temperature change usually requires more energy if no phase change occurs.
  • In phase change calculations, temperature remains constant until the transition is complete.
For this problem, knowing the temperature difference allowed us to use the specific heat capacity to calculate how much energy was needed to heat the tin before it melted.
chemical thermodynamics
Chemical thermodynamics is the broad study of energy transformations in chemical processes, including heat exchanges. This discipline helps us understand how energy is absorbed, released, and conserved in reactions.
In the context of this exercise, chemical thermodynamics is involved in two main parts:
  • The application of specific heat capacity to calculate the energy needed to heat the tin.
  • The use of the heat of fusion to determine how much energy is required to melt the tin once it reaches its melting point.
By combining these properties of tin, we calculate the total energy required to accomplish both tasks, highlighting an important application of chemical thermodynamics in real-world scenarios.
Understanding these calculations provides insights into how energy transformations occur, allowing accurate energy management in heating and melting processes.

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Most popular questions from this chapter

The value of \(\Delta U\) for the decomposition of \(7.647 \mathrm{g}\) of ammonium nitrate can be measured in a bomb calorimeter. The reaction that occurs is \\[ \mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{s}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\] The temperature of the calorimeter, which contains \(415 \mathrm{g}\) of water, increases from \(18.90^{\circ} \mathrm{C}\) to \(20.72^{\circ} \mathrm{C}\) The heat capacity of the bomb is \(155 \mathrm{J} / \mathrm{K}\). What is the value of \(\Delta U\) for this reaction, in \(\mathrm{kJ} / \mathrm{mol}\) ? (IMAGE CANNOT COPY)

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In the lab, you plan to carry out a calorimetry experiment to determine \(\Delta_{r} H\) for the exothermic reaction of \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})\) and \(\mathrm{HCl}(\mathrm{aq}) .\) Predict how each of the following will affect the calculated value of \(\Delta_{t} H .\) (The value calculated for \(\Delta_{i} H\) for this reaction is a negative value so choose your answer from the following: \(\Delta, H\) will be too low [that is, a larger negative valuel, \(\Delta_{r} H\) will be unaffected, \(\Delta_{r} H\) will be too high [that is, a smaller negative value].) (a) You spill a little bit of the \(\mathrm{Ca}(\mathrm{OH})_{2}\) on the benchtop before adding it to the calorimeter. (b) Because of a miscalculation, you add an excess of HCl to the measured amount of \(\mathrm{Ca}(\mathrm{OH})_{2}\) in the calorimeter. (c) \(\mathrm{Ca}(\mathrm{OH})_{2}\) readily absorbs water from the air. The \(\mathrm{Ca}(\mathrm{OH})_{2}\) sample you weighed had been exposed to the air prior to weighing and had absorbed some water. (d) After weighing out \(\mathrm{Ca}(\mathrm{OH})_{2},\) the sample sat in an open beaker and absorbed water. (e) You delay too long in recording the final temperature. (f) The insulation in your coffee-cup calorimeter was poor, so some energy as heat was lost to the surroundings during the experiment. (g) You have ignored the fact that energy as heat also raised the temperature of the stirrer and the thermometer in your system.

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