Problem 91 - Chemistry

Question

Assume that the kinetic energy of a \(1400 \mathrm{~kg}\) car moving at \(115 \mathrm{~km} / \mathrm{h}\) (Problem \(1.90\) ) is converted entirely into heat. How many calories of heat are released, and what amount of water in liters could be heated from \(20.0{ }^{\circ} \mathrm{C}\) to \(50.0^{\circ} \mathrm{C}\) by the car's energy? (One calorie raises the temperature of \(1 \mathrm{~mL}\) of water by \(1{ }^{\circ} \mathrm{C}\).)

Step-by-Step Solution

Answer

5,665 mL (or 5.665 liters) of water can be heated from 20°C to 50°C.

Step 1: Compute Kinetic Energy

To find the kinetic energy of the car, we need to use the formula for kinetic energy: \[ KE = \frac{1}{2}mv^2 \]Where \( m \) is the mass of the car and \( v \) is its velocity. Convert the velocity from km/h to m/s by using the conversion factor \( \frac{1000}{3600} \):\[ v = 115 \times \frac{1000}{3600} \approx 31.94 \text{ m/s} \]Now plug these values into the kinetic energy formula:\[ KE = \frac{1}{2} \times 1400 \times (31.94)^2 \approx 711,307 \text{ J} \]

Step 2: Convert Joules to Calories

Given that 1 Calorie is equal to 4.184 Joules, convert the kinetic energy from Joules to Calories:\[ \text{Calories} = \frac{711307}{4.184} \approx 169,944 \text{ cal} \]

Step 3: Calculate Amount of Water Heated

The problem states that one Calorie raises the temperature of 1 mL of water by 1°C. Calculate how many mL (or liters) of water can be heated from 20°C to 50°C:\[ \Delta T = 50 - 20 = 30^{\circ}C \]The number of Calories required to heat 1 mL of water by 30°C is 30 Calories. Thus, to find the amount of water (in mL) that can be heated from 20°C to 50°C, divide the total Calories by 30:\[ \text{Water Volume} = \frac{169,944}{30} \approx 5,665 \text{ mL} \]Convert this to liters by dividing by 1000:\[ \text{Water Volume} \approx 5.665 \text{ liters} \]

Key Concepts

Calorie to Joule Conversion

Converting energy between different units is a common task in physics, especially when dealing with different types of energy like mechanical and thermal. The calorie is a unit of energy that is commonly used to quantify heat, especially in the contexts of food and chemical reactions. Meanwhile, the joule is the standard unit of energy used in the International System of Units (SI).
To convert calories to joules, we use the conversion factor: \( 1 \text{ cal} = 4.184 \text{ J} \). This means that every calorie of energy is equivalent to 4.184 joules. Conversely, to convert joules to calories, divide the energy in joules by 4.184.Here’s how you can approach such a conversion:

• Identify the amount of energy you have in either joules or calories.
• If converting from joules to calories, divide the joules by 4.184.
• If converting from calories to joules, multiply the calories by 4.184.

In our original exercise, the kinetic energy was 711,307 joules. By dividing this by 4.184, we found that it equaled approximately 169,944 calories.

Heat Transfer

Heat transfer refers to the process by which thermal energy moves from one system or object to another. Understanding heat transfer is essential when dealing with energy changes that involve changes in temperature or state.
This energy movement occurs mainly in three ways:

• Conduction: Direct molecular transfer through solids.
• Convection: Heat transfer through fluids (liquids or gases) as they circulate.
• Radiation: Energy transfer via electromagnetic waves, such as sunlight.

In our case, we’re concerned with heat transfer by conduction, as we are calculating how much heat energy can be transferred to water, causing a rise in temperature. The conversion of kinetic energy into heat indicates how much energy is available to transfer. Knowing the amount of heat transfer helps in calculating how much water can be heated, given a specific temperature change.

Temperature Change

Temperature change is a crucial concept in thermodynamics and is directly related to energy exchange. When we talk about temperature change, we're essentially discussing how the thermal energy in a system alters the temperature of that system.
In terms of water, this relationship is determined by how many calories (or joules) are absorbed or released:

• 1 calorie raises the temperature of 1 mL of water by \(1^{\circ}C\).
• This means, if you know the energy in calories, you can calculate the rise in temperature for a specific volume of water.
  The original problem asked for a temperature increase from \(20^{\circ}C\) to \(50^{\circ}C\), equating to a \(\Delta T\) (temperature change) of \(30^{\circ}C\).
• By understanding the energy available (in calories) and related energy requirements (calories needed to raise each mL by 1°), the number of mL or liters that can be heated can be deduced.

The amount of heat needed per increase in temperature is important when calculating how efficiently energy can be used.

Energy Calculation

Calculating energy involves determining the amount of work done or heat provided to or extracted from a system. In mechanical systems, such as the moving car in our example, kinetic energy can be a source for these calculations.
The kinetic energy \( KE \) of an object is calculated using the formula:\[ KE = \frac{1}{2}mv^2 \]where \( m \) is the mass, and \( v \) is the velocity. With a 1400 kg car moving at 115 km/h converted to 31.94 m/s, the kinetic energy amounts to 711,307 J in our exercise.When this kinetic energy is converted entirely into heat, the total energy available can be used to determine how much water could be heated. By using calorie to joule conversion, the total energy is transformed to calories, revealing how many units of water could be heated across a specified temperature range.
This calculation illustrates how energy can be transitioned between mechanical and thermal forms and helps visualize the practical applications of energy transformations in everyday physics.