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Chapter 6: Problem 61

A 5.00 -g sample of aluminum pellets (specific heat capacity \(=\) 0.89 \(\mathrm{J} / \mathrm{C} \cdot \mathrm{g}\) ) and a \(10.00-\mathrm{g}\) sample of iron pellets (specific heat capacity $=0.45 \mathrm{J} / \mathrm{C} \cdot \mathrm{g}\( are heated to \)100.0^{\circ} \mathrm{C}$ . The mixture of hot iron and aluminum is then dropped into 97.3 \(\mathrm{g}\) water at $22.0^{\circ} \mathrm{C}$ . Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

Short Answer

Expert verified
The final temperature of the metal and water mixture is approximately 25.6°C.

Step by step solution

01

Identify the known values and the unknown final temperature

We are given the following values: - Mass of aluminum, m_aluminum = 5.00 g - Specific heat capacity of aluminum, c_aluminum = 0.89 J/g°C - Mass of iron, m_iron = 10.00 g - Specific heat capacity of iron, c_iron = 0.45 J/g°C - Initial temperature of both aluminum and iron = 100°C - Mass of water, m_water = 97.3 g - Specific heat capacity of water, c_water = 4.18 J/g°C (default value) - Initial temperature of water = 22.0°C The final temperature of the system (water and metals) is unknown. We will call it T_final.
02

Calculate the heat gained by the water

First, we need to calculate the amount of heat gained by the water when it is heated to the final temperature. We can use the formula: q_water = m_water × c_water × (T_final - initial_temperature_water) Note that q_water is positive because it represents the heat gained by the water.
03

Calculate the heat lost by aluminum and iron

Next, we need to calculate the heat lost by the aluminum and iron as they cool down to the final temperature. We can do this for each metal using the formula: q_aluminum = m_aluminum × c_aluminum × (initial_temperature_metal - T_final) q_iron = m_iron × c_iron × (initial_temperature_metal - T_final) Note that q_aluminum and q_iron are both negative because they represent the heat lost by the metals.
04

Apply conservation of energy

Since there is no heat loss to the surroundings, the heat gained by the water must be equal to the heat lost by the metals. Therefore: q_water = q_aluminum + q_iron Now, substitute the values from Steps 2 and 3 and solve for T_final: m_water × c_water × (T_final - initial_temperature_water) = m_aluminum × c_aluminum × (initial_temperature_metal - T_final) + m_iron × c_iron × (initial_temperature_metal - T_final)
05

Solve for T_final

By plugging in the given values, we can now solve for T_final: 97.3 \times 4.18 \times (T_final - 22) = 5.00 \times 0.89 \times (100 - T_final) + 10.00 \times 0.45 \times (100 - T_final) Solving this equation gives us the final temperature: T_final ≈ 25.6°C The final temperature of the metal and water mixture is approximately 25.6°C.

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Most popular questions from this chapter

One of the components of polluted air is NO. It is formed in the high- temperature environment of internal combustion engines by the following reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) \quad \Delta H=180 \mathrm{kJ} $$ Why are high temperatures needed to convert \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) to NO?

What is the difference between \(\Delta H\) and \(\Delta E ?\)

The complete combustion of acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(g),\) produces 1300 . kJ of energy per mole of acetylene consumed. How many grams of acetylene must be burned to produce enough heat to raise the temperature of 1.00 gal water by \(10.0^{\circ} \mathrm{C}\) if the process is 80.0\(\%\) efficient? Assume the density of water is 1.00 \(\mathrm{g} / \mathrm{cm}^{3}\)

If the internal energy of a thermodynamic system is increased by $300 . \mathrm{J}\( while 75 \)\mathrm{J}$ of expansion work is done, how much heat was transferred and in which direction, to or from the system?

Consider 2.00 moles of an ideal gas that are taken from state \(A\) \(\left(P_{A}=2.00 \mathrm{atm}, V_{A}=10.0 \mathrm{L}\right)\) to state \(B\left(P_{B}=1.00 \mathrm{atm}, V_{B}=\right.\) 30.0 \(\mathrm{L}\) ) by two different pathways: These pathways are summarized on the following graph of \(P\) versus \(V :\) Calculate the work (in units of \(\mathrm{J} )\) associated with the two path- ways. Is work a state function? Explain.
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