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Gold has a specific heat of \(0.129 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). When a \(5.00-\mathrm{g}\) piece of gold absorbs \(1.33\) J of heat, what is the change in temperature?

Short Answer

Expert verified
Answer: The change in temperature of the gold piece is 2.06°C.

Step by step solution

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01

Identify the given values

The given values are: - Specific heat of gold (\(c\)) = \(0.129 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) - Mass of the gold piece (\(m\)) = \(5.00\,\text{g}\) - Heat absorbed (\(q\)) = \(1.33\,\text{J}\)
02

Use the formula for heat transfer

We will use the heat transfer formula \(q=mc\Delta T\), where \(q\) is the heat transferred, \(m\) is the mass of the object, \(c\) is the specific heat, and \(\Delta T\) is the change in temperature.
03

Solve for the change in temperature \(\Delta T\)

We will rearrange the equation to solve for the change in temperature: \(\Delta T = \frac{q}{mc}\) Substitute the given values into the equation: \(\Delta T = \frac{1.33\,\text{J}}{5.00\,\text{g} \cdot 0.129 \,\mathrm{J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}}\)
04

Calculate the change in temperature

Now, we will perform the calculation to find the change in temperature: \(\Delta T = \frac{1.33\,\text{J}}{5.00\,\text{g} \cdot 0.129 \,\mathrm{J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}} = \frac{1.33\,\text{J}}{0.645\,\mathrm{J} /{ }^{\circ} \mathrm{C}} = 2.06 { }^{\circ} \mathrm{C}\)
05

State the final answer

The change in temperature of the gold piece when it absorbs \(1.33\,\text{J}\) of heat is \(2.06 { }^{\circ} \mathrm{C}\).

Key Concepts

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

Heat Transfer Formula
Understanding the heat transfer formula is essential in calculating how much energy is needed to change the temperature of a substance. The formula is given by: \[ q = mc\Delta T \] where:
  • \( q \) represents the heat energy transferred (in joules)
  • \( m \) represents the mass of the object (in grams)
  • \( c \) stands for the specific heat capacity (in J/g°C)
  • \( \Delta T \) is the change in temperature (in °C)
The equation shows that the amount of heat energy required depends on the mass of the object, its specific heat capacity, and the temperature change desired.
All three factors impact how easily an object heats up or cools down. In the example, a gold piece is absorbing heat, and we use this formula to find out how its temperature changes.
Temperature Change Calculation
Calculating the change in temperature requires an understanding of how to manipulate the heat transfer formula. If you know the heat energy added or removed from an object, its mass, and specific heat capacity, you can find the temperature change using the equation: \[ \Delta T = \frac{q}{mc} \]By rearranging the heat transfer formula, the temperature change \( \Delta T \) can be isolated for calculation.
Let's break down the calculation for clarity:
  • Determine the heat absorbed, \( q = 1.33 \text{ J} \).
  • Identify the mass of the object, \( m = 5.00 \text{ g} \).
  • Find the specific heat, \( c = 0.129 \text{ J/g°C} \).
  • Substitute these into the formula, \( \Delta T = \frac{1.33 \text{ J}}{5.00 \text{ g} \cdot 0.129 \text{ J/g°C}} \).
  • Perform the calculation: \( \Delta T = \frac{1.33 \text{ J}}{0.645 \text{ J/°C}} = 2.06°C \).
This final result helps you understand that adding 1.33 J of heat to a 5.00 g piece of gold results in a temperature increase of about 2.06°C.
Specific Heat of Gold
The specific heat is an intrinsic property of materials, reflecting how much heat is required to change the temperature of a unit mass by one degree Celsius.
For gold, the specific heat is \( 0.129 \text{ J/g°C} \), meaning it takes \( 0.129 \text{ J} \) to raise the temperature of one gram of gold by one degree Celsius.
This relatively low value indicates that gold heats up quickly with a small energy input.
  • Materials with lower specific heats warm and cool more rapidly compared to those with higher specific heats.
  • Understanding the specific heat helps predict how different materials will respond to heat energy in practical situations.
  • The low specific heat of gold makes it an interesting choice for applications requiring quick thermal changes.
Using the specific heat capacity in our initial calculations, we could determine how much gold's temperature changed after absorbing a specified quantity of heat.

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Most popular questions from this chapter

Given the following reactions, $$ \begin{aligned} \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) & \Delta H^{\circ} &=-534.2 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) & \Delta H^{\circ} &=-241.8 \mathrm{~kJ} \end{aligned} $$ calculate the heat of formation of hydrazine.

When \(35.0 \mathrm{~mL}\) of \(1.43 \mathrm{M} \mathrm{NaOH}\) at \(22.0^{\circ} \mathrm{C}\) is neutralized by \(35.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) also at \(22.0^{\circ} \mathrm{C}\) in a coffee-cup calorimeter, the temperature of the final solution rises to \(31.29^{\circ} \mathrm{C}\). Assume that the specific heat of all solutions is \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), that the density of all solutions is \(1.00 \mathrm{~g} / \mathrm{mL}\), and that volumes are additive. (a) Calculate \(q\) for the reaction. (b) Calculate \(q\) for the neutralization of one mole of \(\mathrm{NaOH}\).

Consider the following reaction in a vessel with a movable piston. $$ \mathrm{X}(g)+\mathrm{Y}(g) \longrightarrow \mathrm{Z}(l) $$ As the reaction occurs, the system loses \(1185 \mathrm{~J}\) of heat. The piston moves down and the surroundings do \(623 \mathrm{~J}\) of work on the system. What is \(\Delta E ?\)

Butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\), is widely used as a fuel for disposable lighters. When one mole of butane is burned in oxygen, carbon dioxide and steam are formed and \(2658.3 \mathrm{~kJ}\) of heat is evolved. (a) Write a thermochemical equation for the reaction. (b) Using Table \(8.3\), calculate the standard heat of formation of butane.

A typical fat in the body is glyceryl trioleate, \(\mathrm{C}_{57} \mathrm{H}_{104} \mathrm{O}_{6}\). When it is metabolized in the body, it combines with oxygen to produce carbon dioxide, water, and \(3.022 \times 10^{4} \mathrm{~kJ}\) of heat per mole of fat. (a) Write a balanced thermochemical equation for the metabolism of fat. (b) How many kilojoules of energy must be evolved in the form of heat if you want to get rid of five pounds of this fat by combustion? (c) How many nutritional calories is this? (1 nutritional calorie = \(1 \times 10^{3}\) calories)

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