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Chapter 17: Problem 39

A copper pot with a mass of 0.500 kg contains 0.170 kg of water, and both are at 20.0\(^\circ\)C. A 0.250-kg block of iron at 85.0\(^\circ\)C is dropped into the pot. Find the final temperature of the system, assuming no heat loss to the surroundings.

Short Answer

Expert verified
The final temperature is approximately 26.5掳C.

Step by step solution

01

Understand the Heat Exchange Concept

In this problem, the heat lost by the iron block will be equal to the heat gained by the pot and the water, because there is no heat loss to the surroundings.
02

Write the Heat Exchange Equations

Use the formula for heat transfer, \[ Q = mc\Delta T \]where \( m \) is mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. Write the equation for each component:1. Iron: \[ Q_{\text{iron}} = m_{\text{iron}} c_{\text{iron}} (T_{\text{final}} - T_{\text{initial}}) \]2. Water: \[ Q_{\text{water}} = m_{\text{water}} c_{\text{water}} (T_{\text{final}} - T_{\text{initial}}) \]3. Copper pot: \[ Q_{\text{copper}} = m_{\text{copper}} c_{\text{copper}} (T_{\text{final}} - T_{\text{initial}}) \]
03

Set Up the Equality Equation

Because the heat lost by the iron is gained by the water and the copper pot, set up the equation as:\[ |Q_{\text{iron}}| = Q_{\text{water}} + Q_{\text{copper}} \]
04

Substitute the Known Values

The specific heat capacities are:- Iron: \( c_{\text{iron}} = 450 \text{ J/(kg}\cdot^\circ\text{C)} \)- Water: \( c_{\text{water}} = 4186 \text{ J/(kg}\cdot^\circ\text{C)} \)- Copper: \( c_{\text{copper}} = 387 \text{ J/(kg}\cdot^\circ\text{C)} \)Substitute these values and the masses into the equations:\[ 0.250 \times 450 \times (T_{\text{final}} - 85) = 0.170 \times 4186 \times (T_{\text{final}} - 20) + 0.500 \times 387 \times (T_{\text{final}} - 20) \]
05

Solve the Equation for Final Temperature

Expand and simplify the equation to solve for \( T_{\text{final}} \):1. \[ 112.5(T_{\text{final}} - 85) = 711.62(T_{\text{final}} - 20) + 193.5(T_{\text{final}} - 20) \]2. Simplify both sides and isolate \( T_{\text{final}} \).3. Solve to find \( T_{\text{final}} \approx 26.5^\circ\text{C} \)

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is a property of a substance that indicates how much energy is needed to change its temperature. It's a crucial factor when studying heat transfer. For any given material, the specific heat capacity (denoted as \( c \)) tells us how much heat (\( Q \)) is required to raise the temperature of a mass \( m \) by a certain change in temperature (\( \Delta T \)). The formula to calculate heat transfer using specific heat capacity is:
  • \( Q = mc\Delta T \)
In our problem, we have three materials involved: iron, water, and copper (from the pot). Each of them has a different specific heat capacity:
  • Iron: 450 J/(kg\cdot^\circ C)
  • Water: 4186 J/(kg\cdot^\circ C)
  • Copper: 387 J/(kg\cdot^\circ C)
This means that water requires more energy to change its temperature compared to copper and iron. Understanding these values allows us to determine how heat will flow between different materials when they are combined.
Thermal Equilibrium
Thermal equilibrium is the state reached when two or more bodies have exchanged heat to the point that there is no further net heat transfer between them. In other words, they've reached the same temperature. In this exercise, an iron block at a high temperature is placed into a cooler copper pot containing water. Initially, there is heat transfer:
  • Iron releases heat as it cools down.
  • The copper pot and water absorb this heat, warming up.
The system reaches thermal equilibrium when the final temperature is the same for all components. To find this equilibrium temperature, we balance the heat lost by the iron with the heat gained by the copper and water. This ensures no heat is lost to the surroundings, allowing us to determine that all the energy stays within the system until equilibrium is achieved.
Calorimetry
Calorimetry is the science of measuring the exchange of heat in physical and chemical processes. This concept is extremely valuable in experiments where temperature changes and heat transfer need to be analyzed. In the given exercise, we use calorimetry principles to determine the final temperature of the system comprising water, iron, and a copper pot. The setup involves:
  • Calculating the heat lost by the iron as it cools down to thermal equilibrium.
  • Calculating the heat gained by the copper pot and the water as they warm up.
  • Setting the heat lost equal to the heat gained because no heat escapes the system.
By measuring initial temperatures, masses, and knowing specific heat capacities, calorimetry lets us solve for the unknown quantity, the final temperature. This problem showcases a typical calorimetry scenario where the final temperature is calculated by aligning the heat exchange in the materials involved.

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Most popular questions from this chapter

A copper calorimeter can with mass 0.446 kg contains 0.0950 kg of ice. The system is initially at 0.0\(^\circ\)C. (a) If 0.0350 kg of steam at 100.0\(^\circ\)C and 1.00 atm pressure is added to the can, what is the final temperature of the calorimeter can and its contents? (b) At the final temperature, how many kilograms are there of ice, how many of liquid water, and how many of steam?

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Convert the following Kelvin temperatures to the Celsius and Fahrenheit scales: (a) the midday temperature at the surface of the moon (400 K); (b) the temperature at the tops of the clouds in the atmosphere of Saturn (95 K); (c) the temperature at the center of the sun \((1.55 \times 10{^7} K)\).

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 gal/min (9.46 L/min) with the tap water being heated from 50\(^\circ\)F (10\(^\circ\)C) to 120\(^\circ\)F (49\(^\circ\)C) 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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