A 275 -g sample of a metal requires \(10.75 \mathrm{~kJ}\) to change its temperature from \(21.2^{\circ} \mathrm{C}\) to its melting temperature, \(327.5^{\circ} \mathrm{C}\). What is the specific heat of this metal?

Short Answer

Expert verified
The specific heat of the metal is approximately 0.1276 J/(g°C).

Step by step solution

01

Identify Given Information

The mass of the metal sample is given as 275 g, the amount of heat transferred is 10.75 kJ, the initial temperature is 21.2°C, and the final temperature is 327.5°C. Note that we need to convert the heat transferred from kJ to J by multiplying by 1000.
02

Convert the Heat Energy to Joules

Since the specific heat capacity formula uses Joules, convert the given energy from kJ to J: Energy in Joules (Q) = 10.75 kJ * 1000 = 10750 J.
03

Calculate the Change in Temperature

The change in temperature (delta T) is the final temperature minus the initial temperature: Delta T = 327.5°C - 21.2°C = 306.3°C.
04

Apply the Formula for Specific Heat

Use the formula for specific heat capacity (c): Q = mc(delta T), where Q is the heat energy in Joules, m is the mass in grams, c is the specific heat capacity, and (delta T) is the change in temperature. Solve for c: c = Q / (m * delta T).
05

Calculate the Specific Heat Capacity

Substitute the given values into the formula and solve for c: c = 10750 J / (275 g * 306.3°C) = 10750 J / (84232.5 g°C) = approximately 0.1276 J/(g°C).

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Key Concepts

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

Heat Transfer
Heat transfer is a cornerstone of thermodynamics that describes the flow of thermal energy due to a temperature difference. Specifically, it occurs from a hotter object to a cooler object until equilibrium is reached. There are three main modes of heat transfer: conduction, which occurs through direct contact; convection, which involves the movement of fluid; and radiation, which transfers energy via electromagnetic waves.

In the context of calculating the specific heat capacity of a metal, heat transfer is directly related to the amount of energy, measured in joules, that the metal must absorb to increase its temperature. In our exercise, a metal sample absorbs 10.75 kJ of energy to increase its temperature from 21.2°C to its melting point at 327.5°C. Understanding heat transfer allows us to determine the amount of heat absorbed or released by the substance when undergoing a temperature change.
Temperature Change
Temperature change refers to the difference in temperature that a substance experiences when exposed to heat transfer. It is a measure of how much the thermal energy of the substance has increased or decreased. The concept of temperature change is essential when calculating specific heat capacity because it determines the amount of heat required to change the temperature of a material by a certain amount.

In our exercise, the temperature change (\( \text{Delta T} \) ) is calculated by subtracting the initial temperature from the final temperature. This change, which is 306.3°C for the metal sample, is used in the specific heat formula to compute the energy needed per gram per degree Celsius (\( \frac{J}{g°C} \) ). Recognizing this change is crucial to comprehend how temperature affects matter and to predict how substances will behave in thermal processes.
Thermodynamics
Thermodynamics is the branch of physics that deals with heat and temperature and their relation to energy and work. It is governed by the laws of thermodynamics, which describe how thermal energy is transferred and transformed in systems.

In this exercise, the first law of thermodynamics, also known as the law of conservation of energy, applies as the heat energy transferred to the metal is used to increase its temperature. No energy is lost or created in the process; it is merely transformed from one form to another. By calculating the specific heat capacity of the metal, we apply thermodynamic principles to identify how efficiently a substance can store thermal energy. This specific heat capacity is a fundamental property that enables scientists and engineers to design systems for heating and cooling, such as radiators, engines, or even cookware, by understanding how different materials react to heat.

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Most popular questions from this chapter

Suppose a ball is sitting at the top of a hill. At this point, the ball has potential energy. The ball rolls down the hill, so the potential energy is converted into kinetic energy. When the ball reaches the bottom of the hill, it goes halfway up the hill on the other side and stops. If energy is supposed to be conserved, then why doesn't the ball go up the other hill to the same level as it started from?

A 110.0-g sample of a gray-colored, unknown, pure metal was heated to \(92.0^{\circ} \mathrm{C}\) and put into a coffee-cup calorimeter containing \(75.0 \mathrm{~g}\) of water at \(21.0^{\circ} \mathrm{C}\). When the heated metal was put into the water, the temperature of the water rose to a final temperature of \(24.2^{\circ} \mathrm{C}\). The specific heat of water is \(4.184 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\). (a) What is the specific heat of the metal? (b) Is it possible that the metal is either iron or lead? Explain.

Indicate with a plus sign \((+)\) any of these processes that require energy and a negative sign \((-)\) any that release energy. (a) riding a bike (d) tires deflating (b) fireworks bursting (e) wood burning in a fireplace (c) water evaporating

At 6:00 P.M., you put a \(300.0\)-g copper pan containing \(800.0 \mathrm{~g}\) of water (all at room temperature, which is \(25^{\circ} \mathrm{C}\) ) on the stove. The stove supplies \(628 \mathrm{~J} / \mathrm{s}\). When will the water reach the boiling point? (Assume no heat is lost.)

A \(225-\mathrm{g}\) sample of aluminum was heated to \(125.5^{\circ} \mathrm{C}\), then placed into \(500.0 \mathrm{~g}\) water at \(22.5^{\circ} \mathrm{C}\). (The specific heat of aluminum is \(0.900 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\) ). Calculate the final temperature of the mixture. (Assume no heat is lost to the surroundings.)

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