/*! 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 65 A 0.250-kg block of a pure mater... [FREE SOLUTION] | 91Ó°ÊÓ

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A 0.250-kg block of a pure material is heated from \(20.0^{\circ} \mathrm{C}\) to \(65.0^{\circ} \mathrm{C}\) by the addition of \(4.35 \mathrm{kJ}\) of energy. Calculate its specific heat and identify the substance of which it is most likely composed.

Short Answer

Expert verified
The block has a specific heat capacity around 387 J/kg°C and is likely composed of brass.

Step by step solution

01

Identify given values

From the problem, we're given: \(m = 0.250 \, \mathrm{kg}\), initial temperature \(T_1 = 20.0^{\circ} \mathrm{C}\), final temperature \(T_2 = 65.0^{\circ} \mathrm{C}\), and \(q = 4.35 \, \mathrm{kJ} = 4350 \, \mathrm{J}\) (because 1kJ = 1000J).
02

Calculate change in temperature

\(\Delta T = T_2 - T_1 = 65.0^{\circ} \mathrm{C} - 20.0^{\circ} \mathrm{C} = 45.0^{\circ} \mathrm{C}\)
03

Determine specific heat

Rearrange the formula \(q=mc\Delta T\) to solve for the specific heat \(c\), yielding \(c = \frac{q}{m \Delta T}\). Substitute the given values to get \(c = \frac{4350 \, \mathrm{J}}{0.250 \, \mathrm{kg} \cdot 45.0^{\circ} \mathrm{C}} \approx 387 \, \mathrm{J/kg \cdot °C}\).
04

Identify the material

The obtained value for \(c\) is closest to the specific heat values of brass or zinc in reference tables. It is necessary to verify with a reliable source to identify the specific substance. For this case, let us assume that it is brass.

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Key Concepts

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

Thermal Energy Transfer
Understanding how thermal energy moves from one object to another, or even within a single object, is essential in the study of thermodynamics. Thermal energy transfer occurs primarily through three mechanisms: conduction, convection, and radiation. In our exercise involving a 0.250-kg block of material, the block absorbs energy and becomes warmer. This process is an example of energy transfer by heat.

Heat is the energy that flows between objects of different temperatures. When heat is added to the material, as indicated by the addition of 4.35 kJ of energy, the thermal energy of the particles within the material increases. This energy increase causes the particles to move more violently and occupy more space, leading to an increase in temperature. The specific heat capacity, represented by the symbol 'c', is a property that tells us how much energy is required to raise the temperature of 1 kg of the substance by 1°C. In this way, specific heat capacity is vital for calculating energy transfer in terms of temperature changes.
Temperature Change
Temperature change is a clear indicator of energy exchange within a system. It signifies the amount of heat absorbed or released by a material in response to heat transfer. The formula for temperature change is \(\Delta T = T_2 - T_1\), where \(T_1\) is the initial temperature and \(T_2\) is the final temperature.

Using the information from the exercise, the initial temperature of the block was \(20.0^{\text{°C}}\), and after absorbing heat, the final temperature reached \(65.0^{\text{°C}}\). The change in temperature \(\Delta T\) is therefore \(45.0^{\text{°C}}\). The importance of calculating \(\Delta T\) lies in its relationship with the thermal energy transferred to the material, which is a key part of the formula used to find the specific heat capacity.
Material Identification
Identifying materials based on their thermal properties is a process utilized in various fields, from material sciences to everyday applications. In the realm of specific heat calculations, once we determine a material's specific heat capacity, we can compare it with known values for different substances to hypothesize the material's identity.

In the conclusion of our exercise, the calculated specific heat of about 387 J/kg·°C is compared against values in a reference table. This comparison suggests that the substance could likely be brass or zinc. Knowledge of specific heat is crucial for material identification as each substance has its unique capacity for storing thermal energy. Reliable data sources are indispensable in confirming the hypothesis, as they contain detailed information on the specific heat capacities of various materials.

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Most popular questions from this chapter

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