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A laboratory technician drops a \(0.0850 \mathrm{~kg}\) sample of unknown solid material, at \(100.0^{\circ} \mathrm{C}\), into a calorimeter. The calorimeter can, initially at \(19.0^{\circ} \mathrm{C},\) is made of \(0.150 \mathrm{~kg}\) of copper and contains \(0.200 \mathrm{~kg}\) of water. The final temperature of the calorimeter can and contents is \(26.1^{\circ} \mathrm{C}\). Compute the specific heat of the sample.

Short Answer

Expert verified
The specific heat of the unknown substance is \(0.994 \) J/g°C.

Step by step solution

01

Find heat gained by water

First find the heat gained by water. The formula to be used is: \(q = mc\Delta T\) where m=0.200 kg, c=4.186 J/g°C is the specific heat of water, and \(\Delta T = 26.1 - 19 = 7.1 \) °C. Thus, \(q_{\mathrm{water}} = 0.200 \times 4.186 \times 7.1 = 5.94426 \) J
02

Find heat gained by calorimeter

Next, find the heat gained by the calorimeter. The specific heat of copper is 0.092 cal/g°C or 0.385 J/g°C. Thus for m=0.150 kg, c= 0.385 J/g°C, and \(\Delta T = 7.1\) °C, \(q_{\mathrm{calorimeter}} = 0.150 \times 0.385 \times 7.1 = 0.411075 \) J
03

Find heat lost by solid

Compute the heat lost by the solid. The heat gained by the water and the calorimeteris equal to the heat lost by the sample. Thus, \(q_{\mathrm{solid}} = q_{\mathrm{water}} + q_{\mathrm{calorimeter}} = 5.94426 + 0.411075 = 6.355335 \) J. Because heat lost is opposite in sign to heat gained, \(q_{\mathrm{solid}} = -6.355335 \) J.
04

Find specific heat of the solid

Lastly, compute the specific heat of the solid. We can express the formula for heat in terms of the specific heat to get \(c = q / (m\Delta T)\). Thus, using m=0.0850 kg and \(\Delta T = 100 - 26.1 = 73.9 \) °C, the specific heat, \(c=\frac{ -6.355335}{0.0850 \times 73.9} = -0.994 \) J/g°C. Since specific heat can't be negative, the actual value is \(0.994 \) J/g°C.

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Key Concepts

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

Calorimetry
Calorimetry is an experimental technique used to measure the amount of heat exchanged in chemical reactions or physical changes. The device used for these measurements is called a calorimeter. The principle behind calorimetry is based on the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred. In the context of the exercise, a laboratory technician is using the calorimeter to find the specific heat of an unknown solid by observing the heat exchange that occurs when the solid is mixed with water.

To perform this, the specific heat capacities of the calorimeter materials (mostly copper) and the water are crucial, as they determine how much the temperature will change when a certain amount of heat is absorbed. The heat gained or lost by a substance can be calculated using the formula:
\[\begin{equation}q = mc\Delta T\end{equation}\] where:
  • \begin{math}q\begin{math} is the amount of heat in joules (J)
  • \begin{math}m\begin{math} is the mass of the substance in kilograms (kg)
  • \begin{math}c\begin{math} is the specific heat capacity in joules per gram degree Celsius (J/g°C)
  • \begin{math}\Delta T\begin{math} is the change in temperature in degrees Celsius (°C)
In this exercise, the objective was to use calorimetry to find the specific heat of an unknown solid based on the heat exchange measured.
Thermal Physics
Thermal physics deals with the study of temperature, heat, and the processes that result from these physical quantities. It encompasses the kinetic theory of gases, which relates the microscopic behaviors of atoms and molecules to the macroscopic properties like temperature and pressure. It also involves the study of heat flow, phase changes, and energy conservation.

In the given exercise, thermal equilibrium is achieved when the solid material, initially at a higher temperature, releases heat to the cooler water and calorimeter until all reach the same final temperature. This is a demonstration of the zeroth law of thermodynamics, which states that if two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other.

This concept is important when reasoning about the final temperature of the mixture in the calorimeter, which was necessary to determine the specific heat of the unknown solid. The calculation required understanding how heat is transferred between objects at different temperatures and how this exchange leads to a new equilibrium state.
Heat Transfer
Heat transfer is the process by which heat energy moves from one object or substance to another. It can occur through different mechanisms: conduction (direct contact transfer), convection (fluid motion transfer), and radiation (electromagnetic waves transfer). In calorimetry, conduction is the primary mode of heat transfer as objects are in direct contact within the calorimeter.

In our step-by-step solution for the unknown solid, we observed that the heat lost by the solid as it cooled down in the calorimeter must equal the heat gained by the water and the calorimeter can. This principle is a specific application of heat transfer and is based on the first law of thermodynamics, which asserts the conservation of energy.

The exercise provided helps in understanding that when a hot object is introduced to cooler surroundings, the heat will flow from the warmer to the cooler substance until thermal equilibrium is reached. The ability to calculate this heat transfer quantitatively is essential in deriving the specific heat of the sample material.

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Most popular questions from this chapter

The molar heat capacity of a certain substance varies with temperature according to the empirical equation $$ C=29.5 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}+\left(8.20 \times 10^{-3} \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}^{2}\right) T $$ How much heat is necessary to change the temperature of \(3.00 \mathrm{~mol}\) of this substance from \(27^{\circ} \mathrm{C}\) to \(227^{\circ} \mathrm{C} ?\) (Hint: Use Eq. (17.18) in the form \(d Q=n C d T\) and integrate. \()\)

On-Demand Water Heaters. Conventional hot-water heaters consist of a tank of water maintained at a fixed temperature. The hot water is to be used when needed. The drawbacks are that energy is wasted because the tank loses heat when it is not in use and that you can run out of hot water if you use too much. Some utility companies are encouraging the use of on-demand water heaters (also known as flash heaters), which consist of heating units to heat the water as you use it. No water tank is involved, so no heat is wasted. A typical household shower flow rate is \(2.5 \mathrm{gal} / \mathrm{min}(9.46 \mathrm{~L} / \mathrm{min})\) with the tap water being heated from \(50^{\circ} \mathrm{F}\left(10^{\circ} \mathrm{C}\right)\) to \(120^{\circ} \mathrm{F}\left(49^{\circ} \mathrm{C}\right)\) by the on-demand heater. What rate of heat input (either electrical or from gas) is required to operate such a unit, assuming that all the heat goes into the water?

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You put a bottle of soft drink in a refrigerator and leave it until its temperature has dropped \(10.0 \mathrm{~K}\). What is its temperature change in (a) \(\mathrm{F}^{\circ}\) and \((\mathrm{b}) \mathrm{C}^{\circ} ?\)

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