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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

\(Camels require very little }}\) water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to \(34.0^{\circ} \mathrm{C}\) overnight and rise to \(40.0^{\circ} \mathrm{C}\) during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a \(400 \mathrm{~kg}\) camel would have to drink if it attempted to keep its body temperature at a constant \(34.0^{\circ} \mathrm{C}\) by evaporation of sweat during the day ( 12 hours) instead of letting it rise to \(40.0^{\circ} \mathrm{C}\). (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, \(3480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). The heat of vaporization of water at \(34^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) )

A copper sphere with density \(8900 \mathrm{~kg} / \mathrm{m}^{3},\) radius \(5.00 \mathrm{~cm}\) and emissivity \(e=1.00\) sits on an insulated stand. The initial temperature of the sphere is \(300 \mathrm{~K}\). The surroundings are very cold, so the rate of absorption of heat by the sphere can be neglected. (a) How long does it take the sphere to cool by \(1.00 \mathrm{~K}\) due to its radiation of heat energy? Neglect the change in heat current as the temperature decreases. (b) To assess the accuracy of the approximation used in part (a), what is the fractional change in the heat current \(H\) when the temperature changes from \(300 \mathrm{~K}\) to \(299 \mathrm{~K} ?\)

In an effort to stay awake for an all-night study session, a student makes a cup of coffee by first placing a \(200 \mathrm{~W}\) electric immersion heater in \(0.320 \mathrm{~kg}\) of water. (a) How much heat must be added to the water to raise its temperature from \(20.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C} ?\) (b) How much time is required? Assume that all of the heater's power goes into heating the water.

In very cold weather a significant mechanism for heat loss by the human body is energy expended in warming the air taken into the lungs with each breath. (a) On a cold winter day when the temperature is \(-20^{\circ} \mathrm{C},\) what amount of heat is needed to warm to body temperature \(\left(37^{\circ} \mathrm{C}\right)\) the \(0.50 \mathrm{~L}\) of air exchanged with each breath? Assume that the specific heat of air is \(1020 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and that \(1.0 \mathrm{~L}\) of air has mass \(1.3 \times 10^{-3} \mathrm{~kg} .\) (b) How much heat is lost per hour if the respiration rate is 20 breaths per minute?

The rate at which radiant energy from the sun reaches the earth's upper atmosphere is about \(1.50 \mathrm{~kW} / \mathrm{m}^{2} .\) The distance from the earth to the sun is \(1.50 \times 10^{11} \mathrm{~m},\) and the radius of the sun is \(6.96 \times 10^{8} \mathrm{~m} .\) (a) What is the rate of radiation of energy per unit area from the sun's surface? (b) If the sun radiates as an ideal blackbody, what is the temperature of its surface?

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