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Chapter 5: Problem 75

You determine that 187 J of energy as heat is required to raise the temperature of \(93.45 \mathrm{g}\) of silver from \(18.5^{\circ} \mathrm{C}\) to \(27.0^{\circ} \mathrm{C} .\) What is the specific heat capacity of silver?

Short Answer

Expert verified
The specific heat capacity of silver is approximately 0.237 J/g掳C.

Step by step solution

01

Understand the Given Values

We have the mass of silver as \(93.45 \text{ g}\), the initial temperature \(18.5^{\circ} \text{C}\), the final temperature \(27.0^{\circ} \text{C}\), and the energy required \(187 \text{ J}\). We need to calculate the specific heat capacity, \(c\), of silver.
02

Determine the Temperature Change

Calculate the change in temperature by subtracting the initial temperature from the final temperature: \[ \Delta T = 27.0^{\circ} \text{C} - 18.5^{\circ} \text{C} = 8.5^{\circ} \text{C} \]
03

Use the Specific Heat Formula

The formula for calculating the specific heat capacity is: \[ q = mc\Delta T \] Where \( q \) is the heat energy, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change. We can rearrange this formula to solve for \( c \): \[ c = \frac{q}{m \Delta T} \]
04

Plug in the Values

Substitute the known values into the rearranged formula:\[ c = \frac{187 \text{ J}}{93.45 \text{ g} \times 8.5^{\circ} \text{C}} \]
05

Calculate the Specific Heat Capacity

Perform the calculation:\[ c = \frac{187}{93.45 \times 8.5} \approx 0.237 \text{ J/g}^{\circ} \text{C} \]
06

Conclusion

The calculated specific heat capacity of silver is approximately \(0.237 \text{ J/g}^{\circ} \text{C}\).

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Key Concepts

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

Energy Transfer
Energy transfer is a fundamental concept in physics, particularly in the context of heat and thermodynamics. When energy is transferred to an object as heat, it causes the particles within that object to move more vigorously. This increased movement is perceived as a rise in temperature. An example of energy transfer is when you hold a cup of hot coffee鈥攜our hand absorbs heat energy from the cup, making it feel warm.
In the exercise provided, 187 joules of energy were transferred to a piece of silver. This transfer changed the silver's temperature from 18.5掳C to 27.0掳C. The amount of energy transferred relates to the heat energy absorbed by the silver to cause its temperature increase. Understanding energy transfer helps us calculate how much heat is needed to change the temperature of a particular substance.
Temperature Change
Temperature change occurs when energy, usually as heat, is added to or removed from a system. This change is measured in degrees Celsius or Fahrenheit, depending on the system used. In the given exercise, the temperature of the silver changed because 187 J of energy was added.
To find the temperature change, you subtract the initial temperature from the final temperature. It's always the difference between where the temperature started and where it ended up. In this case, the silver's temperature increased from 18.5掳C to 27.0掳C, resulting in a temperature change (\(\Delta T\)) of 8.5掳C. Knowing the temperature change is crucial for calculating other properties, such as specific heat capacity.
Heat Energy Calculation
To calculate heat energy required for a temperature change, we use the formula: \[ q = mc\Delta T \]
In this formula:
  • \( q \) represents the heat energy added or removed, measured in joules.
  • \( m \) is the mass of the object, in grams or kilograms.
  • \( c \) is the specific heat capacity, which varies for different substances.
  • \( \Delta T \) is the temperature change.

To determine the specific heat capacity of silver in the example problem, we used the rearranged formula: \[ c = \frac{q}{m \Delta T} \] By inputting the known values: 187 J of energy, a mass of 93.45 g, and a temperature change of 8.5掳C, we calculated:\[ c = \frac{187}{93.45 \times 8.5} \]The calculated specific heat capacity, approximately 0.237 J/g掳C, tells us how much energy is required to raise the temperature of one gram of silver by 1掳C.

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Most popular questions from this chapter

A \(0.692-\mathrm{g}\) sample of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) was burned in a constant-volume calorimeter. The temperature rose from \(21.70^{\circ} \mathrm{C}\) to \(25.22^{\circ} \mathrm{C}\). The calorimeter contained \(575 \mathrm{g}\) of water, and the bomb had a heat capacity of \(650 \mathrm{J} / \mathrm{K}\). What is \(\Delta U\) per mole of glucose?

A piece of titanium metal with a mass of \(20.8 \mathrm{g}\) is heated in boiling water to \(99.5^{\circ} \mathrm{C}\) and then dropped into a coffee-cup calorimeter containing \(75.0 \mathrm{g}\) of water at \(21.7^{\circ} \mathrm{C}\). When thermal equilibrium is reached, the final temperature is \(24.3^{\circ} \mathrm{C}\) Calculate the specific heat capacity of titanium.

The specific heat capacity of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is \(1.74 \mathrm{J} / \mathrm{g} \cdot \mathrm{K} .\) What is its molar heat capacity (in \(\mathrm{J} / \mathrm{mol} \cdot \mathrm{K}) ?\)

A balloon does 324 J of work on the surroundings as it expands under a constant pressure of \(7.33 \times 10^{4} \mathrm{Pa}\). What is the change in volume (in L) of the balloon?

Prepare a graph of specific heat capacities for metals versus their atomic weights. Combine the data in Figure 5.4 and the values in the following table. What is the relationship between specific heat capacity and atomic weight? Use this relationship to predict the specific heat capacity of platinum. The specific heat capacity for platinum is given in the literature as \(0.133 \mathrm{J} / \mathrm{g} \cdot \mathrm{K} .\) How good is the agreement between the predicted and actual values? $$\begin{aligned} &\begin{array}{|l|c|} \hline & \text { Specific Heat Capacity } \\ \text { Metal } & (\mathrm{J} / \mathrm{g} \cdot \mathrm{K}) \\ \hline \text { Chromium } & 0.450 \\ \text { Lead } & 0.127 \\ \text { Silver } & 0.236 \\ \text { Tin } & 0.227 \\ \text { Titanium } & 0.522 \end{array}\\\ &1 \end{aligned}$$
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