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Consider two metals A and B, each having a mass of \(100 \mathrm{~g}\) and an initial temperature of \(20^{\circ} \mathrm{C}\). The specific heat of \(\mathrm{A}\) is larger than that of \(\mathrm{B}\). Under the same heating conditions, which metal would take longer to reach a temperature of \(21^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Metal A will take longer because it has a higher specific heat capacity.

Step by step solution

01

Understanding Specific Heat

The specific heat capacity of a substance dictates how much heat is required to raise the temperature of a unit mass by one degree Celsius. A higher specific heat means more heat is needed to raise the temperature of the substance.
02

Given Data and Requirement

We have two metals A and B, each with a mass of 100 g and an initial temperature of 20°C. The task is to determine which metal will take longer to increase its temperature to 21°C.
03

Heat Required Calculation

The heat required, 'q', to change the temperature of a substance is calculated by the formula: \(q = m imes c imes riangle T\), where \(m\) is the mass, \(c\) is the specific heat, and \(\triangle T\) is the change in temperature.
04

Applying the Formula to Metals A and B

For both metals, the mass (\(m\)) is 100 g, and the temperature change (\(\triangle T\)) is 1°C. However, metal A has a larger specific heat (\(c\)) compared to metal B.
05

Concluding the Result

Since the specific heat of A is larger, more heat is required to change its temperature by 1°C compared to metal B. Thus, under the same heating conditions, metal A will take longer to reach 21°C.

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Key Concepts

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

Heat Required Calculation
To determine how much heat a substance needs to change its temperature, we use a specific formula. This formula is expressed as:
  • \(q = m \times c \times \Delta T\)
Here, \(q\) represents the heat required, \(m\) is the mass of the substance, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.
This relationship shows that the heat required is directly proportional to each factor: mass, specific heat, and temperature change. If any of these quantities increases, the heat needed will also increase.
In our example, both metals A and B have the same mass and experience the same temperature change but differ in specific heat. Therefore, the metal with a larger specific heat will need more energy to achieve the same temperature change.
Temperature Change
Temperature change plays a crucial role in heat calculations. When we talk about temperature change, denoted as \( \Delta T \), it's the difference between the final temperature and the initial temperature. For example, in our metal problem, both metals start at 20°C and go up to 21°C, giving us a temperature change of 1°C.
This change tells us how much energy needs to be absorbed or released by the substance to achieve the desired temperature. A small temperature change would generally mean less heat is needed, all other factors being constant.
Even if two substances have different specific heat capacities, if their temperature change is the same, as in this problem, we just need to focus on how specific heat affects the heat required. This helps manage expectations about how quickly or slowly a substance will heat up or cool down.
Specific Heat Comparison
Specific heat capacity is like a thermal fingerprint for different substances. It tells us how much heat is necessary to raise the temperature of one gram of a substance by one degree Celsius.
In the given exercise, we find out that metal A has a higher specific heat than metal B. This essentially means that metal A can "store" more heat energy for each degree of temperature rise compared to metal B.
With this information, even if they are subjected to the same heating conditions, metal A will require more energy and thus take longer to reach the same temperature increment of 1°C than metal B. This concept helps us understand and compare how materials react to heating or cooling and is crucial in designing materials for specific thermal applications.

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Most popular questions from this chapter

Producer gas (carbon monoxide) is prepared by passing air over red-hot coke: \(\mathrm{C}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g)\) Water gas (a mixture of carbon monoxide and hydrogen) is prepared by passing steam over red-hot coke: \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) For many years, both producer gas and water gas were used as fuels in industry and for domestic cooking. The large-scale preparation of these gases was carried out alternately; that is, first producer gas, then water gas, and so on. Using thermochemical reasoning, explain why this procedure was chosen.

Why are cold, damp air and hot, humid air more uncomfortable than dry air at the same temperatures? [The specific heats of water vapor and air are approximately \(1.9 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) and \(1.0 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) respectively.

A piece of silver with a mass of \(362 \mathrm{~g}\) has a heat capacity of \(85.7 \mathrm{~J} /{ }^{\circ} \mathrm{C}\). What is the specific heat of silver?

A 46-kg person drinks \(500 \mathrm{~g}\) of milk, which has a "caloric" value of approximately \(3.0 \mathrm{~kJ} / \mathrm{g}\). If only 17 percent of the energy in milk is converted to mechanical work, how high (in meters) can the person climb based on this energy intake?

An excess of zinc metal is added to \(50.0 \mathrm{~mL}\) of a \(0.100 \mathrm{M} \mathrm{AgNO}_{3}\) solution in a constant-pressure calorimeter like the one pictured in Figure 5.8 . As a result of the reaction \(\mathrm{Zn}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Ag}(s)\) the temperature rises from \(19.25^{\circ} \mathrm{C}\) to \(22.17^{\circ} \mathrm{C}\). If the heat capacity of the calorimeter is \(98.6 \mathrm{~J} /{ }^{\circ} \mathrm{C},\) calculate the enthalpy change for the given reaction on a molar basis. Assume that the density and specific heat of the solution are the same as those for water, and ignore the specific heats of the metals.

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