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The specific heat of iron metal is \(0.450 \mathrm{~J} / \mathrm{g}-\mathrm{K}\). How many \(J\) of heat are necessary to raise the temperature of a 1.05-kg block of iron from \(25.0^{\circ} \mathrm{C}\) to \(88.5^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The amount of heat necessary to raise the temperature of the 1.05-kg iron block from \(25.0^{\circ} \mathrm{C}\) to \(88.5^{\circ} \mathrm{C}\) is approximately \(29992.5\, \mathrm{J}\).

Step by step solution

01

Understand the formula for heat transfer

The formula for heat transfer, Q, is given by the equation: \[Q = mc\Delta T\] where m is the mass, c is the specific heat, and \(\Delta T\) is the change in temperature.
02

Identify the given values

We are given: - Specific heat (c) of iron metal: \(0.450 \,\mathrm{J/g\cdot K}\) - Mass (m) of the iron block: \(1.05\, \mathrm{kg}\) - Initial temperature (\(T_i\)): \(25.0^\circ \mathrm{C}\) - Final temperature (\(T_f\)): \(88.5^\circ \mathrm{C}\) Before we plug these values into the formula, let's first convert the mass from kg to g: \(1.05 \,\mathrm{kg} \times \frac{1000\, \mathrm{g}}{1\, \mathrm{kg}} = 1050\, \mathrm{g}\) Now, let's calculate the change in temperature (\(\Delta T\)): \(\Delta T = T_f - T_i = 88.5^\circ \mathrm{C} - 25.0^\circ \mathrm{C} = 63.5\, \mathrm{K}\)
03

Calculate the heat transfer

Now that we have all the values, we can plug them into the heat transfer equation: \[Q = (1050\, \mathrm{g}) \times (0.450 \,\mathrm{J/g\cdot K}) \times (63.5\, \mathrm{K})\] Calculate the values: \[Q = 29992.5\, \mathrm{J}\] So, the amount of heat necessary to raise the temperature of the 1.05-kg iron block from \(25.0^{\circ} \mathrm{C}\) to \(88.5^{\circ} \mathrm{C}\) is approximately \(29992.5\, \mathrm{J}\).

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Key Concepts

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

Specific Heat Capacity
When it comes to heating up substances, each material responds differently. This unique response is due to a property called the specific heat capacity, often represented by the symbol 'c'. The specific heat capacity is a measure of how much energy is needed to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin). It's essential to understand that different substances have different specific heat capacities; for example, it takes less energy to raise the temperature of iron than it does for water, which has a high specific heat capacity.

For the iron in our exercise, we see it has a specific heat capacity of 0.450 J/g·K. This means that for every gram of iron, 0.450 joules of energy is needed to increase its temperature by one unit on the Kelvin (or Celsius) scale. When dealing with temperature changes and heat transfer in chemistry, knowing the specific heat capacity of materials is crucial as it helps us predict how a substance will behave when energy is added or removed.

The Role of Specific Heat Capacity in Everyday Life

In our daily lives, the concept of specific heat capacity manifests itself in various ways. For example, when choosing cookware, materials with a lower specific heat capacity like cast iron heat up and cool down quickly, making them ideal for certain types of cooking. Meanwhile, materials like ceramic, with higher specific heat capacities, are better at retaining heat and ensuring a more even cooking process. This principle governs many decisions in product design, environmental science, and engineering.
Temperature Change Calculation
Temperature change, often noted as \(\Delta T\), is the difference in temperature as a substance heats up or cools down. The calculation is simple: subtract the initial temperature (\(T_i\)) from the final temperature (\(T_f\)). This seems straightforward, but the importance of this calculation in thermodynamics can't be overstressed: it allows us to understand how much a substance has warmed or cooled.

Using the right units is also key. In the metric system, temperature is typically recorded in degrees Celsius (°C) or Kelvin (K), and for temperature change, both yield the same value since they're on the same scale – a change of 1°C is equivalent to a change of 1K. This temperature change forms a part of crucial calculations to determine the amount of heat involved in a process.

Applying Temperature Change in the Real World

Daily, we witness temperature change when we boil water, its temperature rises from room temperature to 100°C, indicating a change. Similarly, if we leave a hot beverage to cool, the temperature change tells us how much the beverage has cooled. Accurate temperature change calculations are vital in fields like meteorology for weather predictions, in engineering for material stress-testing, and in cooking for perfecting recipes.
Energy Required for Heating
Determining the energy required to heat a substance is a fundamental concept in chemistry and in practical applications like cooking or industrial processing. To figure out the energy needed, we use the formula \[Q = mc\Delta T\], which combines mass (m), specific heat capacity (c), and temperature change (\Delta T). In the context of our exercise, we calculated the amount of energy necessary to raise the temperature of a 1.05-kg block of iron. Here Q represents the heat in joules – the unit of energy in the International System of Units (SI).

It is important to convert all measurements into compatible units to apply this formula correctly. Notice how we converted the mass of iron from kilograms to grams to match the specific heat capacity units. Once the temperature change and mass are determined, multiplying these values with the substance's specific heat capacity will yield the total energy required.

Understanding Heat Energy in Real-Life

The concept of energy required for heating is evident when boiling water for tea; the stove transfers heat energy into the pot, which then heats the water. On a larger scale, this concept underpins the design of heating systems in buildings, the regulation of temperatures in industrial reactors, and even the determination of nutritional information, as the 'calorie' is a unit of energy originally defined as the amount needed to warm one gram of water by one degree Celsius.

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Most popular questions from this chapter

When a 9.55-g sample of solid sodium hydroxide dissolves in \(100.0 \mathrm{~g}\) of water in a coffee-cup calorimeter (Figure 5.17), the temperature rises from \(23.6^{\circ} \mathrm{C}\) to \(47.4^{\circ} \mathrm{C}\). Calculate \(\Delta H\) (in \(\mathrm{kJ} / \mathrm{mol} \mathrm{NaOH}\) ) for the solution process $$ \mathrm{NaOH}(s) \stackrel{-\cdots}{\mathrm{Na}}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ Assume that the specific heat of the solution is the same as that of pure water.

Consider the decomposition of liquid benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\), to gaseous acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\) : $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \quad \Delta H=+630 \mathrm{~kJ} $$ (a) What is the enthalpy change for the reverse reaction? (b) What is \(\Delta H\) for the formation of 1 mol of acetylene? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If \(C_{6} \mathrm{H}_{6}(g)\) were consumed instead of \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\), would you expect the magnitude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

The enthalpy change for melting ice at \(0{ }^{\circ} \mathrm{C}\) and constant atmospheric pressure is \(6.01 \mathrm{~kJ} / \mathrm{mol}\). Calculate the quantity of energy required to melt a moderately large iceberg with a mass of \(1.25\) million metric tons. (A metric ton is \(1000 \mathrm{~kg}\).)

Identify the force present, and explain whether work is done when (a) a positively charged particle moves in a circle at a fixed distance from a negatively charged particle; (b) an iron nail is pulled off a magnet.

Suppose you toss a tennis ball upward. (a) Does the kinetic energy of the ball increase or decrease as it moves higher? (b) What happens to the potential energy of the ball as it moves higher? (c) If the same amount of energy were imparted to a ball the same size as a tennis ball, but of twice the mass, how high would it go in comparison to the tennis ball? Explain your answers.

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