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Chapter 5: Problem 7

The specific heat capacity of copper metal is \(0.385 \mathrm{J} / \mathrm{g} \cdot \mathrm{K} .\) How much energy is required to heat 168 g of copper from \(-12.2^{\circ} \mathrm{C}\) to \(+25.6^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
2444.74 J

Step by step solution

01

Identify Known Values

We have the specific heat capacity of copper, which is \(c = 0.385 \, \text{J/g} \cdot \text{K}\), the mass \(m = 168 \, \text{g}\), the initial temperature \(T_{i} = -12.2^{\circ} \text{C}\), and the final temperature \(T_{f} = 25.6^{\circ} \text{C}\).
02

Calculate Temperature Change

The change in temperature \( \Delta T \) is calculated using the formula:\[\Delta T = T_{f} - T_{i}\]Substitute the values:\[\Delta T = 25.6^{\circ} \text{C} - (-12.2^{\circ} \text{C}) = 25.6 + 12.2 = 37.8 \, \text{K}\]
03

Utilize the Energy Formula

To find the total energy required, use the specific heat formula:\[Q = m \cdot c \cdot \Delta T\]where \(Q\) is the energy, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the temperature change.
04

Perform the Calculation

Substitute the known values into the formula:\[Q = 168 \, \text{g} \cdot 0.385 \, \text{J/g} \cdot \text{K} \cdot 37.8 \, \text{K}\]Calculate the result:\[Q = 168 \times 0.385 \times 37.8 \]\[ Q = 2444.736 \, \text{J}\]
05

Finalize with Conclusion

Therefore, the total energy required to heat 168 g of copper from \(-12.2^{\circ} \mathrm{C}\) to \(+25.6^{\circ} \mathrm{C}\) is approximately \(2444.74 \, \text{J}\).

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Key Concepts

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

Energy Calculation
Energy calculation is fundamental to understanding how much energy is needed to change the temperature of a substance. When dealing with problems involving heating or cooling, we use a specific formula that considers the mass, specific heat capacity, and temperature change of the material.

The formula for calculating energy is:
  • \[ Q = m \cdot c \cdot \Delta T \]
Where:
  • \( Q \) is the energy in Joules (J)
  • \( m \) is the mass of the substance in grams (g)
  • \( c \) is the specific heat capacity of the substance (J/g路K)
  • \( \Delta T \) is the temperature change in Kelvin (K)
This equation helps calculate the amount of heat energy required when heating a given mass of material by a certain temperature difference. Understanding and effectively utilizing this formula is key to solving thermostat problems like the one concerning copper we examined.
Temperature Change
Temperature change often requires a straightforward calculation to establish how much heat we need to add or subtract. To find the change in temperature, denoted as \( \Delta T \), you subtract the initial temperature from the final temperature.

The equation used is:
  • \[ \Delta T = T_f - T_i \]
Where:
  • \( T_f \) is the final temperature
  • \( T_i \) is the initial temperature
In the case of our copper example, it's important to effectively calculate \( \Delta T \), as this difference determines the amount of energy we need to use or release for temperature change. With correct values, calculating \( \Delta T \) ensures that you input the right numbers into your energy calculation formula.
Copper Specific Heat
Copper's specific heat capacity is a unique property important when performing energy calculations. Specific heat capacity is a measure of how much energy is required to increase the temperature of one gram of a substance by one degree Kelvin (or Celsius).

For copper, this value is known to be:
  • \( c = 0.385 \, \text{J/g} \cdot \text{K} \)
Compared to other substances, copper has a relatively low specific heat capacity, meaning it doesn't require much energy to change its temperature.

This property makes it an excellent material for applications where quick heat exchange is important, such as in electrical components and cookware. Being aware of copper's specific heat capacity allows for precise energy calculations needed for heating or cooling processes, ensuring efficiency in practical applications.

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Most popular questions from this chapter

What quantity of energy, in joules, is required to raise the temperature of \(454 \mathrm{g}\) of tin from room temperature, \(25.0^{\circ} \mathrm{C},\) to its melting point, \(231.9^{\circ} \mathrm{C},\) and then melt the tin at that temperature? (The specific heat capacity of tin is \(0.227 \mathrm{J} / \mathrm{g} \cdot \mathrm{K},\) and the heat of fusion of this metal is \(59.2 \mathrm{J} / \mathrm{g} .\) )

A You have the six pieces of metal listed below, plus a beaker of water containing \(3.00 \times 10^{2} \mathrm{g}\) of water. The water temperature is \(21.00^{\circ} \mathrm{C}.\) $$\begin{array}{|l|c|c|} \hline \text { Metals } & \text { Specific Heat }(J / g K) & \text { Mass }(g) \\\ \hline 1 . A 1 & 0.9002 & 100.0 \\ 2 . A 1 & 0.9002 & 50.0 \\ 3 . A u & 0.1289 & 100.0 \\ 4 . A u & 0.1289 & 50.0 \\ 5 . Z n & 0.3860 & 100.0 \\ 6 . Z n & 0.3860 & 50.0 \\ \hline \end{array}$$ (a) In your first experiment you select one piece of metal and heat it to \(100^{\circ} \mathrm{C},\) and then select a second piece of metal and cool it to \(-10^{\circ} \mathrm{C}\) Both pieces of metal are then placed in the beaker of water and the temperatures equilibrated. You want to select two pieces of metal to use, such that the final temperature of the water is as high as possible. What piece of metal will you heat? What piece of metal will you cool? What is the final temperature of the water? (b) The second experiment is done in the same way as the first. However, your goal now is to cause the temperature to change the least, that is, the final temperature should be as near to \(21.00^{\circ} \mathrm{C}\) as possible. What piece of metal will you heat? What piece of metal will you cool? What is the final temperature of the water?

What does the term standard state mean? What are the standard states of the following substances at \(298 \mathrm{K}: \mathrm{H}_{2} \mathrm{O}, \mathrm{NaCl}, \mathrm{Hg}, \mathrm{CH}_{4} ?\)

Calorimetry Assume you mix 100.0 \(\mathrm{mL}\) of \(0.200 \mathrm{M}\) CsOH with \(50.0 \mathrm{mL}\) of \(0.400 \mathrm{M} \mathrm{HCl}\) in a coffee-cup calorimeter. The following reaction occurs: $$ \mathrm{CsOH}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CsCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) $$ The temperature of both solutions before mixing was \(22.50^{\circ} \mathrm{C},\) and it rises to \(24.28^{\circ} \mathrm{C}\) after the acid-base reaction. What is the enthalpy change for the reaction per mole of CsOH? Assume the densities of the solutions are all \(1.00 \mathrm{g} / \mathrm{mL}\) and the specific heat capacities of the solutions are \(4.2 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}.\)

A 192 -g piece of copper is heated to \(100.0^{\circ} \mathrm{C}\) in a boiling water bath and then dropped into a beaker containing 751 g of water (density = \(1.00 \mathrm{g} / \mathrm{cm}^{3}\) ) at \(4.0^{\circ} \mathrm{C}\). What was the final temperature of the copper and water after thermal equilibrium was reached? \(\left(C_{C u}=0.385 \mathrm{J} / \mathrm{g} \cdot \mathrm{K} .\right).\)
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