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A sheet of gold weighing \(10.0 \mathrm{~g}\) and at a temperature of \(18.0^{\circ} \mathrm{C}\) is placed flat on a sheet of iron weighing \(20.0 \mathrm{~g}\) and at a temperature of \(55.6^{\circ} \mathrm{C}\). What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings.

Short Answer

Expert verified
The final temperature of the combined metals is approximately \(50.9^{\circ}C\).

Step by step solution

01

Understanding the Heat Exchange

In this scenario, since no heat is lost to the surroundings, the heat lost by the sheet of iron will be equal to the heat gained by the sheet of gold. This can be formulated as: \[ q_{ ext{gold}} + q_{ ext{iron}} = 0 \] where \( q \) represents heat.
02

Apply the Heat Transfer Formula

The heat transfer is calculated using the formula: \[ q = mc\Delta T \] where \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature \((T_{ ext{final}} - T_{ ext{initial}})\).
03

Setting Up the Heat Balance Equation

We know the specific heat capacities: - Gold: \( c_{ ext{gold}} = 0.129 \, \text{J/g} \, ^\circ\text{C} \) - Iron: \( c_{ ext{iron}} = 0.450 \, \text{J/g} \, ^\circ\text{C} \) So, the equation becomes: \[ m_{ ext{gold}} c_{ ext{gold}}(T_{ ext{final}} - T_{ ext{gold,initial}}) + m_{ ext{iron}} c_{ ext{iron}}(T_{ ext{final}} - T_{ ext{iron,initial}}) = 0 \] Substituting the values, we get: \[ 10.0 \times 0.129 (T_{ ext{final}} - 18.0) + 20.0 \times 0.450 (T_{ ext{final}} - 55.6) = 0 \]
04

Solve the Equation

First, simplify the equation: \[ 1.29(T_{ ext{final}} - 18.0) + 9.0(T_{ ext{final}} - 55.6) = 0 \] Expanding leads to: \[ 1.29T_{ ext{final}} - 23.22 + 9.0T_{ ext{final}} - 500.4 = 0 \] Combine like terms: \[ 10.29T_{ ext{final}} - 523.62 = 0 \] Solve for \( T_{ ext{final}} \): \[ 10.29T_{ ext{final}} = 523.62 \] \[ T_{ ext{final}} = \frac{523.62}{10.29} \approx 50.9 \]
05

Conclusion: Determine Final Temperature

From the calculation, the final temperature when both metals reach thermal equilibrium is approximately \( 50.9^{ ext{circ}} C \).

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Key Concepts

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

Specific Heat Capacity
Every material requires a certain amount of energy to change its temperature. This is where the concept of specific heat capacity comes in. Specific heat capacity, often denoted as 'c,' is the amount of heat needed to increase the temperature of 1 gram of a substance by 1 degree Celsius.
For instance, gold has a specific heat capacity of 0.129 J/g °C, while iron's is higher at 0.450 J/g °C. This means that iron requires more heat for the same mass to achieve the same temperature change compared to gold.
In our exercise, understanding specific heat capacity helps us calculate how much heat each metal gains or loses as they reach a state of equilibrium.
Thermal Equilibrium
Thermal equilibrium is a key principle in heat transfer. It occurs when two or more bodies, in contact, reach the same temperature and no longer exchange heat.
Think of it as the point where they 'agree' on one, final temperature.
In the exercise, the gold and iron sheets reach thermal equilibrium, which allows us to calculate their shared final temperature by balancing the heat lost and gained between them.
  • The heat lost by the warmer iron is exactly equal to the heat gained by the colder gold.
  • This balance is fundamental in finding the final temperature.
Temperature Change
Temperature change is simply the difference between the final temperature and the initial temperature of a substance. It's represented as ΔT in equations.In heat transfer problems like the exercise, ΔT plays a vital role in determining how much heat is exchanged between the objects.
For each metal:
  • The formula is expressed as \( \Delta T = T_{\text{final}} - T_{\text{initial}} \)
  • For gold, the initial temperature is \(18.0\, ^\circ C\), while iron is \(55.6\, ^\circ C\).
This change helps calculate how much heat is required to reach equilibrium.
Heat Exchange
Heat exchange is the process by which heat energy is transferred between substances or systems. In the scenario with gold and iron, the heat lost by the iron and gained by the gold can be expressed with the formula: \[ q = mc\Delta T \]Here:
  • \( q \) is the heat exchanged;
  • \( m \) is mass;
  • \( c \) is specific heat capacity;
  • \( \Delta T \) is the temperature change.
This equation is used to setup a balance between heat lost by iron and heat gained by gold.
A crucial step is setting up the equation such that all exchanged heat sums to zero. This principle ensures energy is conserved, and by solving the upcoming equation, we can find how the final temperature of the two metals is determined.

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Most popular questions from this chapter

The enthalpy of combustion of benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) is commonly used as the standard for calibrating constant-volume bomb calorimeters; its value has been accurately determined to be \(-3226.7 \mathrm{~kJ} / \mathrm{mol}\). When \(1.9862 \mathrm{~g}\) of benzoic acid are burned in a calorimeter, the temperature rises from \(21.84^{\circ} \mathrm{C}\) to \(25.67^{\circ} \mathrm{C}\). What is the heat capacity of the bomb? (Assume that the quantity of water surrounding the bomb is exactly \(2000 \mathrm{~g} .\) )

Decomposition reactions are usually endothermic, whereas combination reactions are usually exothermic. Give a qualitative explanation for these trends.

Define these terms: enthalpy and enthalpy of reaction. Under what condition is the heat of a reaction equal to the enthalpy change of the same reaction?

A quantity of \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.862 \mathrm{M} \mathrm{HCl}\) is mixed with \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.431 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the \(\mathrm{HCl}\) and \(\mathrm{Ba}(\mathrm{OH})_{2}\) solutions is the same at \(20.48^{\circ} \mathrm{C}\). For the process $$\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$$ the heat of neutralization is \(-56.2 \mathrm{~kJ} / \mathrm{mol}\). What is the final temperature of the mixed solution? Assume the specific heat of the solution is the same as that for pure water.

For reactions in condensed phases (liquids and solids), the difference between \(\Delta H\) and \(\Delta U\) is usually quite small. This statement holds for reactions carried out under atmospheric conditions. For certain geochemical processes, however, the external pressure may be so great that \(\Delta H\) and \(\Delta U\) can differ by a significant amount. A well-known example is the slow conversion of graphite to diamond under Earth's surface. Calculate \(\Delta H-\Delta U\) for the conversion of 1 mole of graphite to 1 mole of diamond at a pressure of 50,000 atm. The densities of graphite and diamond are \(2.25 \mathrm{~g} / \mathrm{cm}^{3}\) and \(3.52 \mathrm{~g} / \mathrm{cm}^{3},\) respectively.

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