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Two solid objects, A and \(\mathrm{B},\) are placed in boiling water and allowed to come to the temperature of the water. Each is then lifted out and placed in separate beakers containing 1000 \(\mathrm{g}\) water at \(10.0^{\circ} \mathrm{C} .\) Object A increases the water temperature by \(3.50^{\circ} \mathrm{C} ; \mathrm{B}\) increases the water temperature by \(2.60^{\circ} \mathrm{C}\) . (a) Which object has the larger heat capacity? (b) What can you say about the specific heats of \(\mathrm{A}\) and \(\mathrm{B} ?\)

Short Answer

Expert verified
Object A has a higher heat capacity as it produces a larger change in water temperature (\(ΔT_A = 3.5^{\circ}C\), \(ΔT_B = 2.6^{\circ}C\)). As for the specific heats, without knowing the objects' masses, we cannot make definitive conclusions. However, if both objects have equal masses, then the specific heat of object A is higher than object B due to its larger heat capacity. If the object with a larger heat capacity has a smaller mass, its specific heat will also be higher.

Step by step solution

01

Identify the Variables Given

We have the following information: - Mass of water in both beakers, m_water = 1000 g - initial water temperature, T_initial = 10.0 °C - final water temperatures, T_final_A = 13.5 °C, T_final_B = 12.6 °C From this, we can calculate the change in temperature for both beakers: ΔT_A = 13.5 °C - 10.0 °C = 3.5 °C ΔT_B = 12.6 °C - 10.0 °C = 2.6 °C
02

Heat Gained by the Water

Let Q_A and Q_B be the amount of heat gained by the water in beakers A and B, respectively. We can calculate these using the given mass, specific heat of water (c_water = 4.18 J/g°C), and change in temperature for each beaker. For object A: \(Q_A = 1000 * 4.18 * 3.5\) For object B: \(Q_B = 1000 * 4.18 * 2.6\)
03

Heat Lost by Objects A and B

Since the heat gained by the water is equal to the heat lost by the objects, we can determine the heat capacities, \(C_A\) and \(C_B\), for objects A and B. \(C_A = Q_A/ΔT_A\) \(C_B = Q_B/ΔT_B\)
04

Compare Heat Capacities

Compare the heat capacities \(C_A\) and \(C_B\) to determine which object has a larger heat capacity. (a) If \(C_A > C_B\), then object A has a larger heat capacity. (b) If \(C_B > C_A\), then object B has a larger heat capacity.
05

Conclusions About Specific Heats

We cannot determine the specific heats of objects A and B without knowing their masses. However, since the heat capacity is directly proportional to specific heat (C = mc), we can make some general observations: - If the object with the larger heat capacity has a smaller mass than the object with the smaller heat capacity, it implies that its specific heat is higher as well. - If both objects have the same mass, the object with the larger heat capacity also has a larger specific heat.

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Key Concepts

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

Specific Heat
Understanding specific heat is key in thermodynamics. It refers to the amount of heat required to change the temperature of one gram of a substance by one degree Celsius. In simple words, it tells us how much energy a material can store. This is why specific heat can vary widely between different materials.

In our exercise, we observed two objects causing different temperature changes in water. While we can infer some information about their specific heats based on their heat transfer characteristics, specific heat calculations require knowing object masses too. Specific heat ties heavily into the heat capacity, as both concepts revolve around energy absorption and temperature change. If a substance has a large specific heat, it will warm up or cool down slowly. Conversely, a smaller specific heat implies faster temperature changes.
Temperature Change
Temperature change is the difference between initial and final temperatures of a substance. In the context of our exercise, we observed two solid objects heated to the temperature of boiling water and then placed in cooler water.

For instance, one object caused a rise of 3.5°C in water while the other increased it by 2.6°C. These changes illustrate how much energy was transferred from each solid object to the surrounding water. Variables like initial temperatures and final temperature measurements play a crucial role in experiments to determine energy transfer efficiency and material properties.

Monitoring temperature shifts provide insights into an object's thermal characteristics and often indicate how heat capacity and specific heat work hand-in-hand during heat transfer.
Heat Transfer
Heat transfer involves the movement of thermal energy from one body to another. In this case, objects A and B lost heat, which was gained by the water, raising its temperature. The principle of heat transfer allows us to determine crucial thermal properties like heat capacity.

Heat transfer is calculated using the formula: \( Q = mc\Delta T \), where \( Q \) is the heat transferred, \( m \) is mass, \( c \) is specific heat, and \( \Delta T \) is the temperature change. In our exercise, knowing the water's specific heat helped calculate the exact heat each water sample absorbed.

Through this understanding, we can compare and contrast the heat capacities of different objects. Object A transferred more heat to the water than object B, indicating its capability to hold and transfer more heat is greater, which could imply a larger heat capacity.

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Most popular questions from this chapter

(a) Write an equation that expresses the first law of thermodynamics in terms of heat and work. (b) Under what conditions will the quantities q and w be negative numbers?

Given the data $$\begin{aligned} \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) & \Delta H=+180.7 \mathrm{kJ} \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) & \Delta H=-113.1 \mathrm{kJ} \\ 2 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & \Delta H=-163.2 \mathrm{kJ} \end{aligned}$$ use Hess's law to calculate \(\Delta H\) for the reaction $$\mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{NO}(g)$$

At the end of \(2012,\) global population was about 7.0 billion people. What mass of glucose in kg would be needed to provide 1500 Cal/person/day of nourishment to the global population for one year? Assume that glucose is metabolized entirely to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) according to the following thermochemical equation: $$\begin{aligned} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) & \\ \Delta H^{\circ}=&-2803 \mathrm{kJ} \end{aligned}$$

We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethylene: $$2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g)$$ Calculate the \(\Delta H^{\circ}\) for this reaction using the following thermochemical data: $$\begin{array}{ll}{\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-890.3 \mathrm{kJ}} \\ {\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)} & {\Delta H^{\circ}=-136.3 \mathrm{kJ}} \\ {2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-571.6 \mathrm{kJ}} \\ {2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-3120.8 \mathrm{kJ}}\end{array}$$

Use bond enthalpies in Table 5.4 to estimate \(\Delta H\) for each of the following reactions: (a) \(\mathrm{H}-\mathrm{H}(g)+\mathrm{Br}-\mathrm{Br}(g) \longrightarrow 2 \mathrm{H}-\mathrm{Br}(g)\) (b)

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