/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 (II) A 0.40-kg iron horseshoe, j... [FREE SOLUTION] | 91Ó°ÊÓ

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(II) A 0.40-kg iron horseshoe, just forged and very hot (Fig. 14\(-\)16), is dropped into 1.25 L of water in a 0.30-kg iron pot initially at 20.0\(^\circ\)C. If the final equilibrium temperature is 25.0\(^\circ\)C, estimate the initial temperature of the hot horseshoe.

Short Answer

Expert verified
The initial temperature of the horseshoe is approximately 174°C.

Step by step solution

01

Identify the Problem Components

The problem involves a hot iron horseshoe, water, and an iron pot reaching thermal equilibrium. We're asked to find the initial temperature of the horseshoe. We know the masses and specific heat capacities of each component, and the final equilibrium temperature.
02

Define Variables and Constants

Let \( T_0 \) be the initial temperature of the horseshoe, \( c_i = 450 \, \text{J/kg°C} \) for iron, and \( c_w = 4186 \, \text{J/kg°C} \) for water. The horseshoe has a mass \( m_h = 0.40 \, \text{kg} \), and the water mass \( m_w = 1.25 \, \text{kg} \). The pot has a mass \( m_p = 0.30 \, \text{kg} \). The initial water and pot temperature is 20.0°C, and the final temperature (equilibrium) is 25.0°C.
03

Use the Principle of Conservation of Energy

The heat lost by the horseshoe equals the heat gained by the water and the pot: \ \( m_h c_i (T_0 - 25) = m_w c_w (25 - 20) + m_p c_i (25 - 20). \)
04

Calculate Heat Gained by Water and Pot

Calculate the heat gained by the water: \ \( Q_w = m_w c_w (25 - 20) = 1.25 \, (4186) \, (5) \). \ Calculate the heat gained by the pot: \ \( Q_p = m_p c_i (25 - 20) = 0.30 \, (450) \, (5) \). \
05

Solve for Initial Temperature of the Horseshoe

The total heat gained by the water and pot is set equal to the heat lost by the horseshoe: \ \( 0.40 \, (450) \, (T_0 - 25) = Q_w + Q_p \). \ Substitute the values from Step 4 and solve for \( T_0 \).
06

Simplify and Calculate

Substitute \( Q_w = 26162.5 \, \text{J} \) and \( Q_p = 675 \, \text{J} \) in Step 5 equation: \ \( 0.40 \, (450) \, (T_0 - 25) = 26162.5 + 675 \). \ Simplify to find \( T_0 \). \ \( 180 (T_0 - 25) = 26837.5 \). \
07

Finalize the Calculation for \( T_0 \)

Simplify the equation: \ \( 180T_0 - 4500 = 26837.5 \). \ Add 4500 to both sides: \ \( 180T_0 = 31337.5 \). \ Divide by 180: \( T_0 = 31337.5 / 180 \). \ Calculate \( T_0 \). \ \( T_0 \approx 174 \text{°C} \).

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Key Concepts

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

Conservation of Energy
In the context of solving the problem involving a hot iron horseshoe, water, and an iron pot, the principle of conservation of energy is fundamental. Energy cannot be created or destroyed—only transferred. By understanding this, we can solve problems that involve multiple objects reaching thermal equilibrium.
The conservation of energy tells us that the total energy within a closed system remains constant. When the horseshoe is dropped into the water, it transfers heat energy until both the horseshoe and the water reach the same temperature. This is what we call thermal equilibrium.
  • Heat lost by the horseshoe equals the heat gained by the water and the pot.
  • The total heat transfer in and out of the system equals zero, illustrating energy conservation.
We use these principles to create equations that describe the energy balance in the system. These equations then help us find unknown quantities, like the initial temperature of the horseshoe.
Specific Heat Capacity
Specific heat capacity is a property that determines how much heat energy is required to change the temperature of a substance by a certain amount. Every material has a unique specific heat capacity, which tells us how it interacts with heat.
In our exercise here, we deal with iron and water, two substances with vastly different specific heat capacities:
  • Iron has a specific heat capacity of 450 J/kg°C, meaning it takes 450 joules to raise 1 kg of iron by 1°C.
  • Water, on the other hand, needs 4186 joules per kg per degree Celsius for the same temperature change.
Understanding specific heat capacity helps to determine how different materials will react when heat energy is added or removed.
In our problem, knowing the specific heat capacities allows us to calculate the amount of heat transferred to or from each material when the horseshoe and water interact until thermal equilibrium is achieved.
Calorimetry
Calorimetry is the science of measuring the amount of heat in chemical reactions or physical changes. It involves using the principles of heat transfer to determine energy changes in a system.
In practical terms, calorimetry allows us to calculate how much heat is exchanged when objects at different temperatures interact. For instance, in our problem, calorimetry principles are used to find the initial temperature of the hot horseshoe when it is placed in cooler water.
  • The calorimetry equation used incorporates the masses, specific heat capacities, and temperature changes of the horseshoe, water, and pot.
  • By setting the heat lost by the horseshoe equal to the heat gained by the water and the pot, we can solve for the unknown initial temperature using the calorimetry equation.
Thus, calorimetry serves as a practical tool for understanding and calculating thermal processes in real-world applications. It is invaluable in solving problems involving heat exchange and energy conservation.

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Most popular questions from this chapter

(II) A 28-g ice cube at its melting point is dropped into an insulated container of liquid nitrogen. How much nitrogen evaporates if it is at its boiling point of 77 K and has a latent heat of vaporization of 200 kJ/kg? Assume for simplicity that the specific heat of ice is a constant and is equal to its value near its melting point.

(II) The \(heat \, capacity\), \(C\), of an object is defined as the amount of heat needed to raise its temperature by 1 C\(^\circ\). Thus, to raise the temperature by requires heat \(Q\) given by \(\Delta T\) $$Q = C \, \Delta \, T$$ (\(a\)) Write the heat capacity C in terms of the specific heat, \(c\), of the material. (\(b\)) What is the heat capacity of 1.0 kg of water? (\(c\)) Of 45 kg of water?

A soft-drink can contains about 0.35 kg of liquid at 5\(^\circ\)C. Drinking this liquid can actually consume some of the fat in the body, since energy is needed to warm the liquid to body temperature (37\(^\circ\)C). How many food Calories should the drink have so that it is in perfect balance with the heat needed to warm the liquid (essentially water)?

(II) High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbed from your body if you: (\(a\)) eat 1.0 kg of \(-15^\circ\)C snow which your body warms to body temperature of 37\(^\circ\)C; (\(b\)) melt 1.0 kg of \(-15^\circ\)C snow using a stove and drink the resulting 1.0 kg of water at 2\(^\circ\)C, which your body has to warm to 37\(^\circ\)C.

(II) Determine the latent heat of fusion of mercury using the following calorimeter data: 1.00 kg of solid Hg at its melting point of \(-39.0^\circ\)C is placed in a 0.620-kg aluminum calorimeter with 0.400 kg of water at 12.80\(^\circ\)C; the resulting equilibrium temperature is 5.06\(^\circ\)C.

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