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Chapter 8: Problem 2

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

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 \(2.06 { }^{\circ} \mathrm{C}\).

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Most popular questions from this chapter

On a hot day, you take a six-pack of soda on a picnic, cooling it with ice. Each empty (aluminum) can weighs \(12.5 \mathrm{~g}\) and contains \(12.0\) oz of soda. The specific heat of aluminum is \(0.902 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\); take that of soda to be \(4.10 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) (a) How much heat must be absorbed from the six-pack to lower the temperature from \(25.0^{\circ}\) to \(5.0^{\circ} \mathrm{C}\) ? (b) How much ice must be melted to absorb this amount of heat? \(\left(\Delta H_{\mathrm{fus}}\right.\) of ise is given in Table 8.2.)

When one mole of caffeine \(\left(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}\right)\) is burned in air, \(4.96 \times 10^{3} \mathrm{~kJ}\) of heat is evolved. Five grams of caffeine is burned in a bomb calorimeter. The temperature is observed to increase by \(11.37^{\circ} \mathrm{C}\). What is the heat capacity of the calorimeter in \(\mathrm{J} /{ }^{\circ} \mathrm{C} ?\)

Nitroglycerine, \(\mathrm{C}_{3} \mathrm{H}_{5}\left(\mathrm{NO}_{3}\right)_{3}(l)\), is an explosive most often used in mine or quarry blasting. It is a powerful explosive because four gases \(\left(\mathrm{N}_{2}\right)\) \(\mathrm{O}_{2}, \mathrm{CO}_{2}\), and steam) are formed when nitroglycerine is detonated. In addition, \(6.26 \mathrm{~kJ}\) of heat is given off per gram of nitroglycerine detonated. (a) Write a balanced thermochemical equation for the reaction. (b) What is \(\Delta H\) when \(4.65\) mol of products is formed?

Mercury has a specific heat of \(0.140 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). Assume that a thermometer has 20 (2 significant figures) grams of mercury. How much heat is absorbed by the mercury when the temperature in the thermometer increases from \(98.6^{\circ} \mathrm{F}\) to \(103.2^{\circ} \mathrm{F} ?\) (Assume no heat loss to the glass of the thermometer.)

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.
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