Question

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.)

Short answer

Answer: The mercury absorbs approximately 8.1 J of heat.

Step-by-step solution

Convert temperatures from Fahrenheit to Celsius

To convert the temperatures from Fahrenheit to Celsius, we can use the formula: \(C = \frac{5}{9}(F - 32)\) For the initial temperature \(98.6^{\circ} F\), the conversion will be as follows: $$ C_{initial} = \frac{5}{9}(98.6 - 32) $$ For the final temperature \(103.2^{\circ} F\), the conversion will be as follows: $$ C_{final} = \frac{5}{9}(103.2 - 32) $$

Calculate the change in temperature

Now we need to find the change in temperature (\(\Delta T\)) which is the difference between the final and initial temperatures in Celsius. $$ \Delta T = C_{final} - C_{initial} $$

Find the heat absorbed by the mercury

To find the heat absorbed by the mercury, we will use the formula: \(Q = mc\Delta T\) Where: \(Q\) = heat absorbed (Joules) \(m\) = mass of the mercury (grams) \(c\) = specific heat capacity of mercury (J/g℃) \(\Delta T\) = change in temperature (℃) Given: \(m\) = \(20 g\) (2 significant figures) \(c\) = \(0.140 J/g\cdot℃\) Substitute the values into the formula and calculate the heat absorbed: $$ Q = 20 \times 0.140 \times \Delta T $$ Now, calculate the answer and provide the final result with the appropriate significant figures.

Key concepts

Heat Absorption

Understanding how substances absorb heat is fundamental in thermodynamics. Heat absorption refers to the process of a substance taking in energy in the form of heat, causing its temperature to increase if there is no phase change (like melting or boiling).

The amount of heat absorbed is directly proportional to the mass of the substance—more mass means potentially more heat absorbed. It also depends on the substance's specific heat capacity, which is a measure of how much energy is needed to raise the temperature of one gram of the substance by one degree Celsius. Specific heat capacity is unique to every substance; for example, water has a high specific heat capacity, making it great for cooling systems.

When solving problems related to heat absorption, we use the formula:
\[Q = mc\Delta T\]
where
\(Q\) represents the heat absorbed (measured in Joules, J),
\(m\) stands for the mass of the substance (in grams, g),
\(c\) is the specific heat capacity (J/g°C), and
\(\Delta T\) denotes the change in temperature (in degrees Celsius).

To apply this concept correctly, it is crucial to ensure that the temperature change is expressed in Celsius and that the specific heat capacity is aligned with the substance in question.

Temperature Conversion

When working with heat-related exercises, it's often necessary to convert temperatures from one scale to another. The most common scales used are Celsius (°C) and Fahrenheit (°F). To make accurate calculations, particularly in heat absorption problems, temperature conversion is an essential skill.

The formula to convert Fahrenheit to Celsius is:
\[ C = \frac{5}{9}(F - 32) \]
This transformation is required because specific heat capacities are typically provided in Celsius units, and many scientific calculations are based on the Celsius or Kelvin scales.

Let's consider the thermometer mercury example. The initial temperature in Fahrenheit is converted to Celsius to use it in the specific heat formula. Both the initial and final temperatures must be converted to make certain that \(\Delta T\), the change in temperature, is correctly calculated as the difference between the final and initial temperatures in degrees Celsius.

Significant Figures

Significant figures are a way of expressing precision in numerical answers. They are the digits in a number that carry meaning contributing to its measurement resolution. This concept is particularly relevant when calculating figures in scientific contexts, such as heat absorption, where precision is paramount.

For example, in the thermometer problem, the mass of mercury is given as 20 grams, which has two significant figures. This means that the final answer should also reflect two significant figures to maintain consistency in precision.

Applying significant figures correctly involves a few rules:

• All non-zero digits are significant.
• Zeros between non-zero digits are significant.
• Leading zeros are not significant.
• Trailing zeros are significant if there's a decimal point.

When multiplying or dividing, the number of significant figures in the result should match the number with the least significant figures used in the calculation. In our heat equation, after computing the heat absorbed, we must round off the answer to reflect the same level of precision as provided in the given data.