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Chapter 12: Problem 44

A piece of glass has a temperature of \(83.0^{\circ} \mathrm{C}\). Liquid that has a temperature of \(43.0^{\circ} \mathrm{C}\) is poured over the glass, completely covering it, and the temperature at equilibrium is \(53.0^{\circ}\) C. The mass of the glass and the liquid is the same. Ignoring the container that holds the glass and liquid and assuming that the heat lost to or gained from the surroundings is negligible, determine the specific heat capacity of the liquid.

Short Answer

Expert verified
The specific heat capacity of the liquid is 2.52 J/g掳C.

Step by step solution

01

Define the Known Variables

We need to identify the important given variables for this problem. \(T_\text{initial, glass} = 83.0^\circ C\), \(T_\text{initial, liquid} = 43.0^\circ C\), \(T_\text{equilibrium} = 53.0^\circ C\), and the mass of the glass \(m_g = m_l\) (same as the mass of the liquid).
02

Express the Heat Transfer Equations

The heat lost by the glass \(Q_g = m_g c_g (T_\text{initial, glass} - T_\text{equilibrium})\) and the heat gained by the liquid \(Q_l = m_l c_l (T_\text{equilibrium} - T_\text{initial, liquid})\). Since the mass is the same, use \(m\) for both.
03

Use the Conservation of Energy Principle

Since no heat is lost to the surroundings, the heat lost by the glass is equal to the heat gained by the liquid: \[Q_g = Q_l\]\[m c_g (T_\text{initial, glass} - T_\text{equilibrium}) = m c_l (T_\text{equilibrium} - T_\text{initial, liquid})\].
04

Substitute Known Values and Simplify

Substitute the known values: \[c_g (83.0 - 53.0) = c_l (53.0 - 43.0)\]\[c_g \times 30.0 = c_l \times 10.0\].
05

Insert the Specific Heat Capacity of Glass and Solve

The specific heat capacity of glass is \(c_g = 0.84\, \text{J/g掳C}\). Replace \(c_g\) in the equation: \[0.84 \times 30.0 = c_l \times 10.0\]\[25.2 = 10 c_l\].
06

Solve for the Specific Heat Capacity of the Liquid

Divide both sides by 10 to find \(c_l\): \[c_l = \frac{25.2}{10} = 2.52\, \text{J/g掳C}\].

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Key Concepts

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

Heat Transfer
Heat transfer is the process of thermal energy moving from a hotter object to a cooler one. This phenomenon occurs due to the temperature difference between the two entities. In our example, the glass is warmer at the start, while the liquid is cooler.
  • As soon as these two items come into contact, heat begins to transfer from the glass to the liquid.
  • The ultimate goal is to reach a balance, where both objects attain the same temperature, known as thermal equilibrium.
The heat transfer continues until there's no longer a temperature difference, thus achieving equilibrium.
Energy Conservation
The principle of energy conservation states that energy in an isolated system cannot be created or destroyed. In the context of calorimetry, this means that heat lost by one material must be equal to the heat gained by another.
  • In the given problem, heat lost by the glass is equal to the heat gained by the liquid since no heat is lost to the surroundings.
  • This ensures that total energy is conserved throughout the process.
Mathematically, it means:\[m c_g (T_{\text{initial, glass}} - T_{\text{equilibrium}}) = m c_l (T_{\text{equilibrium}} - T_{\text{initial, liquid}})\]
Thermal Equilibrium
Thermal equilibrium is reached when two substances within a closed system reach the same temperature. At this point, no heat flows between them because the temperature is uniform. In the example provided, the glass and liquid reach a common final temperature of \(53.0^{\circ}C\).
  • This new temperature indicates that the system has reached thermal equilibrium.
  • The lack of additional heat flow suggests that energy distribution between the glass and the liquid is balanced.
Understanding thermal equilibrium allows scientists to make calculations about the specific heat capacities of various materials.
Calorimetry
Calorimetry is the science of measuring the amount of heat involved in a chemical or physical process. By observing changes in temperature, we can infer energy exchanges and understand the properties of substances.
  • In the exercise, calorimetry techniques help us determine the specific heat capacity of the liquid in question.
  • By applying the concept of energy conservation and knowing the specific heat capacity of the glass, we can solve for the unknown specific heat capacity of the liquid.
This is done using the formula:\[c_l = \frac{c_g \times (T_{\text{initial, glass}} - T_{\text{equilibrium}})}{(T_{\text{equilibrium}} - T_{\text{initial, liquid}})}\]Calorimetry helps in understanding heat transfer in chemical reactions and changes in matter.

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Most popular questions from this chapter

A \(0.200-\mathrm{kg}\) piece of aluminum that has a temperature of \(-155^{\circ} \mathrm{C}\) is added to \(1.5 \mathrm{~kg}\) of water that has a temperature of \(3.0^{\circ} \mathrm{C}\). At equilibrium the temperature is \(0.0^{\circ} \mathrm{C}\). Ignoring the container and assuming that the heat exchanged with the surroundings is negligible, determine the mass of water that has been frozen into ice.

Two grams of liquid water are at \(0{ }^{\circ} \mathrm{C}\), and another two grams are at \(100{ }^{\circ} \mathrm{C}\). Heat is removed from the water at \(0{ }^{\circ} \mathrm{C}\), completely freezing it at \(0{ }^{\circ} \mathrm{C}\). This heat is then used to vaporize some of the water at \(100{ }^{\circ} \mathrm{C}\). What is the mass (in grams) of the liquid water that remains?

The latent heat of vaporization of \(\mathrm{H}_{2} \mathrm{O}\) at body temperature \(\left(37.0^{\circ} \mathrm{C}\right)\) is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) To cool the body of a \(75-\mathrm{kg}\) jogger [average specific heat capacity \(\left.=3500 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\right]\) by \(1.5 \mathrm{C}^{\circ},\) how many kilograms of water in the form of sweat have to be evaporated?

A steel ruler is calibrated to read true at \(20.0^{\circ} \mathrm{C}\). A draftsman uses the ruler at \(40.0^{\circ} \mathrm{C}\) to draw a line on a \(40.0^{\circ} \mathrm{C}\) copper plate. As indicated on the warm ruler, the length of the line is \(0.50 \mathrm{~m}\). To what temperature should the plate be cooled, such that the length of the line truly becomes \(0.50 \mathrm{~m}\) ?

An unknown material has a normal melting/ freezing point of \(-25.0^{\circ} \mathrm{C},\) and the liquid phase has a specific heat capacity of \(160 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\). One-tenth of a kilogram of the solid at \(-25.0^{\circ} \mathrm{C}\) is put into a \(0.150-\mathrm{kg}\) aluminum calorimeter cup that contains 0.100 kg of glycerin. The temperature of the cup and the glycerin is initially \(27.0^{\circ} \mathrm{C}\). All the unknown material melts, and the final temperature at equilibrium is \(20.0^{\circ} \mathrm{C}\). The calorimeter neither loses energy to nor gains energy from the external environment. What is the latent heat of fusion of the unknown material?
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