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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
48244.17 J

Step by step solution

01

Calculate the Energy Needed to Raise Temperature

First, calculate the energy required to raise the temperature of tin from its initial temperature to its melting point using the formula: \[ Q_1 = m \cdot c \cdot \Delta T \]where \( Q_1 \) is the energy needed for heating, \( m = 454 \ \mathrm{g} \) is the mass of the tin, \( c = 0.227 \ \mathrm{J/g \cdot K} \) is the specific heat capacity of tin, and \( \Delta T = 231.9^{\circ} \mathrm{C} - 25.0^{\circ} \mathrm{C} = 206.9^{\circ} \mathrm{C} \) is the temperature change. Substitute the values into the formula: \[ Q_1 = 454 \cdot 0.227 \cdot 206.9 = 21367.37 \ \mathrm{J} \]
02

Calculate the Energy Needed for Melting

Next, calculate the energy required to melt the tin using the formula: \[ Q_2 = m \cdot L_f \]where \( Q_2 \) is the energy needed for melting, \( m = 454 \ \mathrm{g} \) is the mass of the tin, and \( L_f = 59.2 \ \mathrm{J/g} \) is the heat of fusion. Substitute the values into the formula: \[ Q_2 = 454 \cdot 59.2 = 26876.8 \ \mathrm{J} \]
03

Calculate Total Energy Required

Finally, sum the energy calculated in Steps 1 and 2 to find the total energy required:\[ Q_{\text{total}} = Q_1 + Q_2 = 21367.37 + 26876.8 = 48244.17 \ \mathrm{J} \]
04

Answer

The total energy required to raise the temperature of 454 g of tin from room temperature to its melting point and to melt it is approximately 48244.17 J.

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Key Concepts

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

Heat of Fusion
When substances transform from solid to liquid, such as tin melting, they require a specific amount of energy called the heat of fusion. This energy is crucial because it breaks the bonds holding the solid structure together, turning it into a liquid. In the exercise, we focus on how much energy tin needs to reach this change at its melting point, which is distinctively different from the energy needed for heating. The heat of fusion for a particular material is its fingerprint for this phase change.

With tin, the heat of fusion is given as 59.2 J/g. This means for every gram of tin, 59.2 Joules of energy is necessary to transform it into liquid form once it hits 231.9 °C. To calculate the energy for melting all the tin, multiply the mass (in grams) by this heat of fusion.

This calculated energy merely covers the phase change at the melting point, not the temperature hike up to it. Getting comfortable with this concept is key in many science and engineering problems.
Temperature Change
Understanding how temperature change affects a substance is essential in thermodynamics. When we increase the temperature of a material, it requires energy, and the amount necessary depends on the substance's specific heat capacity.

Specific heat capacity refers to how much energy is needed to change the temperature of one gram of a substance by 1 °C (or 1 K). For tin, this is 0.227 J/g·K. Before melting, tin first needs to be heated from 25.0 °C to 231.9 °C, which represents a temperature change (ΔT) of 206.9 °C.

When calculating how much energy tin requires to rise to its melting point, use the formula \(Q_1 = m \cdot c \cdot \Delta T\). The exercise shows that using this formula, the required energy amounts to 21367.37 J. Such calculations ensure we account for all energy changes before reaching the melting point.
Energy Calculation
In any thermodynamics exercise, finding out the total energy required for a process involves breaking it into smaller components. In our example, after dealing with the heat necessary for raising the temperature and for the phase change, the final step is to combine these energy calculations.

The total energy needed is the sum of the energy to raise the temperature (\(Q_1\)) and the energy for melting the tin (\(Q_2\)).

All energy contributions add up: \[Q_{\text{total}} = Q_1 + Q_2 = 21367.37 + 26876.8 = 48244.17 \ \mathrm{J}\]

Understanding these steps clarifies how substances behave under different conditions and how much energy these transitions require. It's a straightforward approach: divide the task into heating and phase change, calculate each separately, and then combine for the total energy requirement.

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Most popular questions from this chapter

A piece of lead with a mass of \(27.3 \mathrm{g}\) was heated to \(98.90^{\circ} \mathrm{C}\) and then dropped into \(15.0 \mathrm{g}\) of water at \(22.50^{\circ} \mathrm{C} .\) The final temperature was \(26.32^{\circ} \mathrm{C} .\) Calculate the specific heat capacity of lead from these data.

A 192 -g piece of copper is heated to \(100.0^{\circ} \mathrm{C}\) in a boiling water bath and then dropped into a beaker containing \(751 \mathrm{g}\) of water (density \(=1.00 \mathrm{g} / \mathrm{cm}^{3}\) ) at \(4.0^{\circ} \mathrm{C} .\) What was the final temperature of the copper and water after thermal equilibrium was reached? \(\left(C_{\mathrm{Cu}}=0.385 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\right)\)

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 CAN'T COPY)

Suppose you burned \(0.300 \mathrm{g}\) of \(\mathrm{C}(\mathrm{s})\) in an excess of \(\mathrm{O}_{2}(\mathrm{g})\) in a constant volume calorimeter to give \(\mathrm{CO}_{2}(\mathrm{g})\) \(\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})\) The temperature of the calorimeter, which contained 775 g of water, increased from \(25.00^{\circ} \mathrm{C}\) to \(27.38^{\circ} \mathrm{C}\) The heat capacity of the bomb is \(893 \mathrm{J} / \mathrm{K}\). Calculate \(\Delta U\) per mole of carbon.

Insoluble \(\mathrm{PbBr}_{2}(\mathrm{s})\) precipitates when solutions of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\) and \(\mathrm{NaBr}(\mathrm{aq})\) are mixed. \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NaBr}(\mathrm{aq}) \rightarrow \mathrm{PbBr}_{2}(\mathrm{s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})\) $$ \Delta_{\mathrm{r}} H^{\circ}=? $$ To measure the enthalpy change, \(200 .\) mL of \(0.75 \mathrm{M}\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\) and \(200 . \mathrm{mL}\) of \(1.5 \mathrm{M} \mathrm{NaBr}(\mathrm{aq})\) are mixed in a coffee-cup calorimeter. The temperature of the mixture rises by \(2.44^{\circ} \mathrm{C} .\) Calculate the enthalpy change for the precipitation of \(\mathrm{PbBr}_{2}(\mathrm{s}),\) in \(\mathrm{k} \mathrm{J} / \mathrm{mol}\). (Assume the density of the solution is \(1.0 \mathrm{g} / \mathrm{mL},\) and its specific heat capacity is \(4.2 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\).

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