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How many calories are required to heat each of the following from \(15^{\circ} \mathrm{C}\) to \(65^{\circ} \mathrm{C} ?(\) a) \(3.0\) g of aluminum, \((b) 5.0 \mathrm{~g}\) of Pyrex glass, (c) 20 g of platinum. The specific heats, in \(\mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), for aluminum, Pyrex, and platinum are \(0.21,0.20\), and 0.032, respectively.

Short Answer

Expert verified
(a) 31.5 cal, (b) 50 cal, (c) 32 cal.

Step by step solution

01

Understand the formula

To calculate the calories needed to heat a substance, use the formula: \[ Q = mc\Delta T \]where \( Q \) is the heat energy (calories), \( m \) is the mass in grams, \( c \) is the specific heat capacity in cal/g°C, and \( \Delta T \) is the change in temperature in °C.
02

Calculate change in temperature (ΔT)

The change in temperature \( \Delta T \) is the difference between the final temperature and the initial temperature. For every part (a, b, c) of this problem: \( \Delta T = 65^{\circ} \text{C} - 15^{\circ} \text{C} = 50^{\circ} \text{C} \).
03

Calculate calories for aluminum (a)

For aluminum, the mass \( m = 3.0 \text{ g} \) and the specific heat \( c = 0.21 \text{ cal/g°C} \). Substitute these values into the formula:\[ Q = (3.0 \text{ g}) \times (0.21 \text{ cal/g°C}) \times (50^{\circ} \text{C}) \]\[ Q = 31.5 \text{ cal} \].
04

Calculate calories for Pyrex glass (b)

For Pyrex glass, the mass \( m = 5.0 \text{ g} \) and the specific heat \( c = 0.20 \text{ cal/g°C} \). Substitute these values into the formula:\[ Q = (5.0 \text{ g}) \times (0.20 \text{ cal/g°C}) \times (50^{\circ} \text{C}) \]\[ Q = 50 \text{ cal} \].
05

Calculate calories for platinum (c)

For platinum, the mass \( m = 20 \text{ g} \) and the specific heat \( c = 0.032 \text{ cal/g°C} \). Substitute these values into the formula:\[ Q = (20 \text{ g}) \times (0.032 \text{ cal/g°C}) \times (50^{\circ} \text{C}) \]\[ Q = 32 \text{ cal} \].

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Key Concepts

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

Thermochemistry
Thermochemistry is a branch of chemistry that deals with the study of energy changes, particularly heat, during chemical reactions and physical transformations. It helps us understand how energy is absorbed or released, aiding in the development of processes that require controlled energy input. By examining how substances interact with heat, we can optimize industrial processes, improve energy efficiency, and innovate new technologies. Understanding the flow of energy and how substances transform during phase changes or chemical reactions is central to thermochemistry. For example, knowing how much energy is required to heat a substance is crucial for tasks ranging from designing engines to cooking food. This discipline provides the quantitative tools needed to calculate these heat changes accurately.
Heat Transfer Calculations
Heat transfer calculations are essential when we need to determine the amount of energy required to change the temperature of a substance. These calculations use specific formulas that consider the mass of the substance, its specific heat capacity, and the temperature change desired. This is often approached using the formula:
  • \[ Q = mc\Delta T \]
  • \( Q \): Heat energy required (in calories or joules)
  • \( m \): Mass of the substance (in grams)
  • \( c \): Specific heat capacity (the amount of heat needed to change the temperature of one gram of the substance by one degree Celsius)
  • \( \Delta T \): Temperature change (in degrees Celsius)
By using this formula, we can effectively calculate the amount of energy needed for heating or cooling processes, crucial for practical applications in everyday life and various industries.
Temperature Change
Temperature change, denoted as \( \Delta T \), is a simple yet fundamental concept in heat calculations. It represents the difference between the final and initial temperatures of a substance. To understand temperature change better:
  • Calculate \( \Delta T \) by subtracting the initial temperature from the final temperature.
  • In our example, if the material starts at \( 15^{\circ} \text{C} \) and is heated to \( 65^{\circ} \text{C} \), the temperature change is:\[ \Delta T = 65^{\circ} \text{C} - 15^{\circ} \text{C} = 50^{\circ} \text{C} \]
Knowing the temperature change is critical because it directly influences how much energy is required to reach the desired temperature. Accurately determining \( \Delta T \) ensures precise heat transfer calculations.
Calorimetry
Calorimetry is a technique used to measure the amount of heat absorbed or released during chemical reactions and phase changes. It forms the experimental backbone of thermochemistry, allowing scientists to quantify energy changes.
  • Calorimeters are devices used to perform these measurements, consisting of an insulated container that helps prevent heat exchange with the environment.
  • By measuring the heat transfer involved in a process, calorimetry enables the calculation of enthalpy changes, specific heat capacities, and more.
  • This technique is vital in various fields including chemistry, biology, and material science, where it is used to study thermophysical properties and understand reaction energetics.
Practically, calorimeters help in determining how much energy is supplied during heating or cooling processes, aiding in accurate heat transfer calculations.
Heat Energy Formula
The heat energy formula \( Q = mc\Delta T \) is fundamental for calculating the energy required to change the temperature of a given mass of substance. This formula bridges the gap between theoretical concepts and real-world applications in thermochemistry.- **Variables Explanation**:
  • \( Q \) denotes the total heat energy (in calories or joules) that is absorbed or released.
  • \( m \) is the mass of the substance involved, given in grams.
  • \( c \) represents the specific heat capacity, indicating how much heat per unit mass is needed to raise the temperature by one degree Celsius.
  • \( \Delta T \) is the difference between the initial and final temperature.
By understanding and applying this formula, students and professionals can predict the energy requirements for thermal processes, ensuring accurate and efficient control over heating or cooling systems.

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Most popular questions from this chapter

How long does it take a 2.50-W heater to boil away \(400 \mathrm{~g}\) of liquid helium at the temperature of its boiling point \((4.2 \mathrm{~K})\) ? For helium, \(L_{v}=5.0 \mathrm{cal} / \mathrm{g}\).

Calculate the heat of fusion of ice from the following data for ice at \(0{ }^{\circ} \mathrm{C}\) added to water: \(\begin{array}{lr}\text { Mass of calorimeter } & 60 \mathrm{~g} \\ \text { Mass of calorimeter plus water } & 460 \mathrm{~g} \\ \text { Mass of calorimeter plus water and ice } & 618 \mathrm{~g} \\ \text { Initial temperature of water } & 38.0^{\circ} \mathrm{C} \\ \text { Final temperature of mixture } & 5.0^{\circ} \mathrm{C} \\ \text { Specific heat of calorimeter } & 0.10 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\end{array}\)

Suppose a 60-kg person consumes 2500 Cal of food in one day. If the entire heat equivalent of this food were retained by the person's body, how large a temperature change would it cause? (For the body, \(c=0.83 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\).) Remember that \(1 \mathrm{Cal}=1 \mathrm{kcal}=\) 1000 cal. The equivalent amount of heat added to the body in one day is $$ \Delta Q=(2500 \text { Cal })(1000 \mathrm{cal} / \mathrm{Cal})=2.5 \times 10^{6} \mathrm{cal} $$ Then, by use of \(\Delta Q=m c \Delta T\), $$ \Delta T=\frac{\Delta Q}{m c}=\frac{2.5 \times 10^{6} \mathrm{cal}}{\left(60 \times 10^{3} \mathrm{~g}\right)\left(0.83 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)}=50^{\circ} \mathrm{C} $$

Suppose a person who eats 2500 Cal of food each day loses the heat equivalent of the food through evaporation of water from the body. How much water must evaporate each day? At body temperature, \(L_{v}\) for water is about \(600 \mathrm{cal} / \mathrm{g}\).

Victoria Falls on the Zambezi River is \(108 \mathrm{~m}\) high, and \(1088 \mathrm{~m}^{3}\) of water pours over it every second. Assuming no loss in energy, what is the rise in temperature of the water due to the drop? [Hint: Think PE.

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