Question

A 30.5 -g sample of an alloy at \(93.0^{\circ} \mathrm{C}\) is placed into \(50.0 \mathrm{~g}\) of water at \(22.0^{\circ} \mathrm{C}\) in an insulated coffee-cup calorimeter with a heat capacity of \(9.2 \mathrm{~J} / \mathrm{K}\). If the final temperature of the system is \(31.1^{\circ} \mathrm{C},\) what is the specific heat capacity of the alloy?

Short answer

The specific heat capacity of the alloy is approximately 1.05 J/g·K.

Step-by-step solution

- Identify the variables

Determine the mass, initial temperature, and final temperature for both the alloy and the water, as well as the heat capacity for the calorimeter.\[ m_{alloy} = 30.5 \text{ g} \] \[ T_{initial, alloy} = 93.0^{\text{C}} \] \[ m_{water} = 50.0 \text{ g} \] \[ T_{initial, water} = 22.0^{\text{C}} \] \[ T_{final} = 31.1^{\text{C}} \] \[ C_{calorimeter} = 9.2 \text{ J/K} \]

- Calculate the heat gained by the water

Use the formula for heat gained or lost: \(q = mc\triangle T\). For water: \[ q_{water} = m_{water} \times c_{water} \times (T_{final} - T_{initial, water}) \] Here, \(c_{water} = 4.18 \text{ J/g}\bullet \text{K} \). Thus, \[ q_{water} = 50.0 \text{ g} \times 4.18 \text{ J/g}\bullet K \times (31.1 - 22.0)^{\text{C}} \] \[ q_{water} = 50.0 \times 4.18 \times 9.1 \] \[ q_{water} = 1903.9 \text{ J} \]

- Calculate the heat absorbed by the calorimeter

Use the formula: \( q_{calorimeter} = C_{calorimeter} \times \triangle T \) So, \[ q_{calorimeter} = 9.2 \text{ J/K} \times (31.1 - 22.0) \text{K} \] \[ q_{calorimeter} = 9.2 \times 9.1 \] \[ q_{calorimeter} = 83.72 \text{ J} \]

- Calculate the total heat gained by the system

Add the heats gained by water and the calorimeter: \[ q_{\text{total}} = q_{water} + q_{calorimeter} \] \[ q_{\text{total}} = 1903.9 \text{ J} + 83.72 \text{ J} \] \[ q_{\text{total}} = 1987.62 \text{ J} \]

- Calculate the heat lost by the alloy

Since the heat gained by the system is equal to the heat lost by the alloy: \[ q_{\text{alloy}} = - q_{\text{total}} \] \[ q_{\text{alloy}} = -1987.62 \text{ J} \]

- Calculate the specific heat capacity of the alloy

Rearrange the formula \( q = mc\triangle T \) to find the specific heat capacity: \[ c = \frac{-q}{m \times \triangle T} \] So, for the alloy: \[ c_{\text{alloy}} = \frac{1987.62 \text{ J}}{30.5 \text{ g} \times (93.0 - 31.1)^{\text{C}}} \] \[ c_{\text{alloy}} = \frac{1987.62}{30.5 \times 61.9} \] \[ c_{\text{alloy}} = \frac{1987.62}{1887.95} \approx 1.05 \text{ J/g}\bullet\text{K} \]

Key concepts

specific heat capacity

Specific heat capacity is the amount of heat needed to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin). It's a property that varies from material to material. In our exercise, we had to find the specific heat capacity of an alloy. By using the formula \(c = \frac{-q}{m \times \triangle T}\), we calculated it as 1.05 J/g·K. Understanding specific heat capacity helps in identifying how different substances react to heat changes and thereby predict how much energy is required for a certain temperature change.

heat transfer

Heat transfer is the process of moving heat from one substance to another. In the given exercise, heat was transferred from a hot alloy to cooler water until thermal equilibrium was reached. This concept is fundamental in thermodynamics. There are three primary ways heat can be transferred:

• Conduction: Direct transfer of heat through a material.
• Convection: Transfer of heat by the movement of fluids.
• Radiation: Transfer of heat through electromagnetic waves.

Understanding heat transfer is crucial in explaining how heat moves in various systems and how energy balance is maintained.

temperature change

Temperature change occurs when heat is added or removed from a substance. In the exercise, the temperature of the water increased from 22.0°C to 31.1°C because heat was transferred from the alloy. We can link temperature change to heat transfer using the formula \(q = mc\triangle T\), where \(q\) is heat, \(m\) is mass, \(c\) is specific heat capacity, and \(\triangle T\) is the change in temperature. Calculating temperature change is essential in thermodynamics for finding out how much heat energy is involved in heating or cooling a substance.

thermodynamics

Thermodynamics is the study of heat, energy, and the work done by them. It involves principles that explain how energy moves and changes states. Some of the core laws of thermodynamics are:

• First Law: Energy cannot be created or destroyed, only transferred or converted from one form to another. This principle was used to equalize the heat gained by water and the heat lost by the alloy.
• Second Law: Heat naturally flows from a hotter object to a cooler one. This explains why the alloy cooled down while the water warmed up.
• Third Law: As temperature approaches absolute zero, the entropy of a perfect crystal approaches a constant minimum.

These basic laws help us understand how and why energy changes occur in different systems.