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A \(50.0-\mathrm{g}\) sample of metal at \(100.00{ }^{\circ} \mathrm{C}\) is added to \(60.0 \mathrm{~g}\) water that is initially \(25.00{ }^{\circ} \mathrm{C}\). The final temperature of both the water and the metal is \(31.51^{\circ} \mathrm{C}\). (a) Use the specific heat of water to find the heat absorbed by the water. (b) How much heat did the metal sample lose? (c) Calculate the specific heat of the metal.

Short Answer

Expert verified
(a) Heat absorbed by water: 1634.228 J. (b) Heat lost by metal: -1634.228 J. (c) Specific heat of metal: 0.477 J/g°C.

Step by step solution

01

Determine Heat Absorbed by Water

The heat absorbed by the water can be calculated using the formula:\[ q = m \cdot c \cdot \Delta T \]where \( m = 60.0 \text{ g} \) is the mass of the water, \( c = 4.18 \text{ J/g°C} \) is the specific heat of water, and \( \Delta T \) is the change in temperature of the water. Calculate \( \Delta T \):\[ \Delta T = T_{\text{final}} - T_{\text{initial}} = 31.51^{\circ} \text{C} - 25.00^{\circ} \text{C} = 6.51^{\circ} \text{C} \]Now, substitute the values:\[ q = 60.0 \times 4.18 \times 6.51 \approx 1634.228 \text{ J} \]
02

Calculate Heat Lost by Metal

According to the law of conservation of energy, the heat lost by the metal is equal to the heat gained by the water (assuming no heat is lost to the surroundings). Therefore, the heat lost by the metal is:\[ q_{\text{metal}} = -q_{\text{water}} \]\[ q_{\text{metal}} = -1634.228 \text{ J} \] (The negative sign indicates energy loss from the metal.)
03

Determine Specific Heat of the Metal

To find the specific heat \( c_{\text{metal}} \) of the metal, use the heat formula:\[ q = m \cdot c \cdot \Delta T \]Rearrange to solve for \( c \):\[ c_{\text{metal}} = \frac{q}{m \cdot \Delta T} \]The mass of the metal \( m = 50.0 \text{ g} \) and its temperature change \( \Delta T = T_{\text{initial}} - T_{\text{final}} = 100.00^{\circ} \text{C} - 31.51^{\circ} \text{C} = 68.49^{\circ} \text{C} \).Substitute the known values:\[ c_{\text{metal}} = \frac{-1634.228}{50.0 \cdot 68.49} \approx -0.477 \text{ J/g°C} \](We take the absolute value of specific heat, so \( c_{\text{metal}} = 0.477 \text{ J/g°C} \).)

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Key Concepts

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

Heat Absorption
When you add a hot object to a cooler one, heat flows from the hot to the cool item. This heat transfer is called 'heat absorption' by the cooler substance. In our exercise, we have water absorbing heat from a metal sample. To calculate the heat absorbed by the water, we use the formula:\[ q = m \cdot c \cdot \Delta T \]Here \( q \) stands for the heat absorbed, \( m \) is the mass of the substance—in this case, the water, \( c \) denotes the specific heat capacity, and \( \Delta T \) signifies the change in temperature.
  • Mass of water (\( m \)): 60.0 g
  • Specific heat of water (\( c \)): 4.18 J/g°C
  • Temperature change (\( \Delta T \)): \( 31.51^{\circ} \text{C} - 25.00^{\circ} \text{C} = 6.51^{\circ} \text{C} \)
Substituting these values gives us the heat absorbed. This simple calculation is crucial, as it sets the stage for understanding how heat is transferred between substances.
Temperature Change
Temperature change, denoted by \( \Delta T \), is the difference between the final temperature and the initial temperature of a substance. It helps us understand how much a substance's temperature changes due to heat transfer. In this exercise, both water and metal undergo temperature changes.For the water:
  • Initial Temperature: \( 25.00^{\circ} \text{C} \)
  • Final Temperature: \( 31.51^{\circ} \text{C} \)
  • \( \Delta T \) for water: \( 6.51^{\circ} \text{C} \)
For the metal:
  • Initial Temperature: \( 100.00^{\circ} \text{C} \)
  • Final Temperature: \( 31.51^{\circ} \text{C} \)
  • \( \Delta T \) for metal: \( 68.49^{\circ} \text{C} \)
Understanding these temperature differences is essential for accurate calculations and sets the groundwork for calculating specific heat values.
Conservation of Energy
The principle of conservation of energy states that energy cannot be created or destroyed. This principle comes into play during heat transfer between substances. In our example, the heat lost by the metal (a hot object) is equal to the heat gained by the water (a cooler object).
  • Heat lost by metal: \(-1634.228 \text{ J}\)
  • Heat gained by water: \(+1634.228 \text{ J}\)
The negative sign for the metal indicates it is losing energy, while a positive sign for the water indicates energy gained. This equality is crucial because it allows us to use the heat absorbed or lost to find specific heat capacities. Calculations often assume no heat is lost to surroundings, keeping the process "closed" for simplicity.
Calorimetry
Calorimetry is a technique used to measure the amount of heat transferred to or from a substance. It's like taking your activity's temperature but on an energy scale! In calorimetry, you measure the change in temperature to determine how much heat is absorbed or lost. This exercise is an example of calorimetry. By carefully measuring temperatures before and after mixing the metal with water, we can calculate:
  • The heat absorbed by water
  • The heat lost by the metal
  • The specific heat of the metal
Such methods are foundational in chemistry and physics for understanding thermal processes. They enable us to quantify what occurs when objects of different temperatures come into contact. By mastering calorimetry, we gain insight into energy interactions in everyday life. So next time you witness steam from a hot beverage cooling down, you'll know it’s a real-life calorimetry lesson!

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Most popular questions from this chapter

The thermite reaction produces a large quantity of heat, enough to melt the iron metal that is a product of the reaction: $$ \begin{aligned} 2 \mathrm{Al}(\mathrm{s})+\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s}) \rightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~s})+2 \mathrm{Fe}(\mathrm{s}) \\ \Delta H_{\mathrm{rxn}} &=-852 \mathrm{~kJ} \end{aligned} $$ What is the enthalpy change if \(50.0 \mathrm{~g}\) Al reacts with excess iron(III) oxide?

Methane, \(\mathrm{CH}_{4}(\mathrm{~g}),\) and octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(\ell),\) are important components of the widely used fossil fuels. The enthalpy change for combustion of \(1 \mathrm{~mol}\) methane is \(-890 \mathrm{~kJ},\) and that for 1 mol octane is \(-5466 \mathrm{~kJ} .\) Which of these fuels produces more energy per gram of compound burned? What is the difference in energy produced per gram of compound?

Why is it unnecessary to include the enthalpies of formation of elements, such as \(\mathrm{P}_{4}(\mathrm{~s}), \mathrm{H}_{2}(\mathrm{~g})\), or \(\mathrm{C}\) (graphite), in a table of standard enthalpies of formation?

Addition of solid ammonium nitrate to water in a coffeecup calorimeter results in a solution with a temperature lower than the original temperature of the water. The \(\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{~s})\) absorbs heat in the process of dissolving, $$ \mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{~s})+\text { heat } \rightarrow \mathrm{NH}_{4}^{+}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) $$ If the calorimeter is perfectly insulating (no heat can enter or leave), what provides the heat?

When \(7.11 \mathrm{~g} \mathrm{NH}_{4} \mathrm{NO}_{3}\) is added to \(100 \mathrm{~mL}\) water, the temperature of the calorimeter contents decreases from \(22.1^{\circ} \mathrm{C}\) to \(17.1^{\circ} \mathrm{C}\). Assuming that the mixture has the same specific heat as water and a mass of \(107 \mathrm{~g}\), calculate the heat \(q\). Is the dissolution of ammonium nitrate exothermic or endothermic?

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