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Chapter 5: Problem 31

A piece of titanium metal with a mass of \(20.8 \mathrm{g}\) is heated in boiling water to \(99.5^{\circ} \mathrm{C}\) and then dropped into a coffee-cup calorimeter containing \(75.0 \mathrm{g}\) of water at \(21.7^{\circ} \mathrm{C} .\) When thermal equilibrium is reached, the final temperature is \(24.3^{\circ} \mathrm{C} .\) Calculate the specific heat capacity of titanium.

Short Answer

Expert verified
The specific heat capacity of titanium is approximately 0.52 J/g掳C.

Step by step solution

01

Identify known values

First, let's list what we know from the problem: - Mass of titanium ( mt ) = 20.8 g - Initial temperature of titanium ( T_{i,t} ) = 99.5 掳C - Mass of water ( mw ) = 75.0 g - Initial temperature of water ( T_{i,w} ) = 21.7 掳C - Final temperature of both systems ( T_{f} ) = 24.3 掳C - Specific heat capacity of water ( cw ) = 4.18 J/g掳C.
02

Write the energy balance equation

Since energy is conserved, the heat lost by titanium equals the heat gained by water. Thus, we can write:\[m_t c_t (T_{i,t} - T_f) = m_w c_w (T_f - T_{i,w})\]where mt is the mass of titanium, c_t is the specific heat capacity of titanium, T_{i,t} is the initial temperature of titanium, T_f is the final temperature, T_{i,w} is the initial temperature of water, m_w is the mass of water, and c_w is the specific heat capacity of water.
03

Substitute known values and solve for specific heat capacity

Insert the known values into the equation:\[20.8 \, \text{g} \cdot c_t \cdot (99.5 - 24.3) = 75.0 \, \text{g} \cdot 4.18 \, \frac{\text{J}}{\text{g} \, ^\circ \text{C}} \cdot (24.3 - 21.7)\]Calculate both sides,- Left side: \[20.8 \, \text{g} \cdot c_t \cdot 75.2\]- Right side: \[75.0 \, \text{g} \cdot 4.18 \, \frac{\text{J}}{\text{g} \, ^\circ \text{C}} \cdot 2.6^\circ \text{C} = 815.4 \, \text{J}\]Set the equations equal and solve for c_t :\[20.8 \, \text{g} \cdot c_t \cdot 75.2 = 815.4\] \[c_t = \frac{815.4}{20.8 \cdot 75.2}\]Calculate c_t to find the specific heat capacity of titanium.
04

Calculate and interpret result

Perform the final calculation:\[c_t = \frac{815.4}{1565.76}\approx 0.52 \, \text{J/g掳C}\]This means that the specific heat capacity of titanium is approximately 0.52 J/g掳C.

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Key Concepts

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

Thermal Equilibrium
Thermal equilibrium is a vital concept in understanding how heat exchange works between different objects. When two objects at different temperatures come into contact, they will exchange heat energy until they reach the same temperature. This shared temperature is known as the thermal equilibrium. In the given exercise, titanium and water equilibrate to a final temperature of \(24.3^{\circ} \text{C}\).
The energy lost by the hotter titanium is equal to the energy gained by the cooler water. This is because no heat is lost to the surroundings in an ideal calorimetry scenario. Once both substances reach the same temperature, heat transfer ceases. Thermal equilibrium helps us predict the final temperature and set up equations needed for calculating specific heat values.
Calorimetry
Calorimetry is the science of measuring the change in heat associated with chemical reactions or physical changes. In this problem, a coffee-cup calorimeter is used to measure heat exchange between titanium and water. Calorimeters help us by isolating the process to ensure accurate heat transfer measurement without external heat loss or gain.
In the exercise, we determine the specific heat of titanium by monitoring how it exchanges heat with water, whose specific heat is known. Calorimetry allows us to apply equations that represent the amount of heat transferred, connecting the masses, specific heats, and temperature changes of the substances involved.
  • Mass and specific heat: These are crucial for determining how much heat an object can hold.
  • Temperature changes: They indicate the amount of heat transfer involved in reaching thermal equilibrium.
Through careful measurement and proper setup, calorimetry provides a precise way to calculate specific heat capacities, such as that of titanium in this exercise.
Energy Conservation
Energy conservation is a fundamental concept in physics, stating that energy cannot be created or destroyed, only transformed. When dealing with calorimetry, this principle is crucial. The heat lost by one material must be equal to the heat gained by another when they exchange energy.
In the problem, energy conservation is mathematically expressed as the equivalence between the heat loss of titanium and the heat gain of water. Setting these two quantities equal allows us to solve for the unknown specific heat capacity of titanium.
  • Identify known variables such as mass, specific heat, and temperatures.
  • Use the energy balance equation to equate the heat transfer.
  • Solve for the specific heat of the unknown material.
This method underscores the power of energy conservation in predicting and calculating the behavior of materials during temperature change.

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Most popular questions from this chapter

The molar heat capacity of mercury is \(28.1 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\) What is the specific heat capacity of this metal in \(\mathrm{J} / \mathrm{g} \cdot \mathrm{K}\) ?

When 0.850 g of Mg was burned in oxygen in a constant volume calorimeter, \(25.4 \mathrm{kJ}\) of energy as heat was evolved. The calorimeter was in an insulated container with \(750 . \mathrm{g}\) of water at an initial temperature of \(18 . \overline{6}^{\circ} \mathrm{C}\). The heat capacity of the bomb in the calorimeter is \(820 . \mathrm{J} / \mathrm{K}\) (a) Calculate \(\Delta U\) for the oxidation of \(\mathrm{Mg}\) (in \(\mathrm{k} \mathrm{J} / \mathrm{mol}\) \(\mathrm{Mg})\) (b) What will be the final temperature of the water and the bomb calorimeter in this experiment?

You wish to know the enthalpy change for the formation of liquid \(\mathrm{PCl}_{3}\) from the elements. $$ \mathrm{P}_{4}(\mathrm{s})+6 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{3}(\ell) \quad \Delta_{\mathrm{r}} H^{\circ}=? $$ The enthalpy change for the formation of \(\mathrm{PCl}_{5}\) from the elements can be determined experimentally, as can the enthalpy change for the reaction of \(\mathrm{PCl}_{3}(\ell)\) with more chlorine to give \(\mathrm{PCl}_{5}(\mathrm{s}):\) \(\begin{aligned} \mathrm{P}_{4}(\mathrm{s})+10 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{5}(\mathrm{s}) & \\ \Delta_{r} H^{\circ} &=-1774.0 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} \\\ \mathrm{PCl}_{3}(\ell)+\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{PCl}_{5}(\mathrm{s}) & \\ \Delta_{\mathrm{r}} H^{\circ} &=-123.8 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} \end{aligned}\) Use these data to calculate the enthalpy change for the formation of 1.00 mol of \(\mathrm{PCl}_{3}(\ell)\) from phosphorus and chlorine.

A piece of lead with a mass of \(27.3 \mathrm{g}\) was heated to \(98.90^{\circ} \mathrm{C}\) and then dropped into \(15.0 \mathrm{g}\) of water at \(22.50^{\circ} \mathrm{C} .\) The final temperature was \(26.32^{\circ} \mathrm{C} .\) Calculate the specific heat capacity of lead from these data.

You are attending summer school and living in a very old dormitory. The day is oppressively hot, there is no air-conditioner, and you can't open the windows of your room. There is a refrigerator in the room, however. In a stroke of genius, you open the door of the refrigerator, and cool air cascades out. The relief does not last long, though. Soon the refrigerator motor and condenser begin to run, and not long thereafter the room is hotter than it was before. Why did the room warm up?
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