Question

Which gives up more heat on cooling from \(50^{\circ} \mathrm{C}\) to \(10^{\circ} \mathrm{C}\) \(50.0 \mathrm{g}\) of water or \(100 .\) g of ethanol (specific heat capacity of ethanol \(=2.46 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}) ?\)

Short answer

Ethanol gives up more heat, losing 9840 J compared to water's 8360 J.

Step-by-step solution

Understand the Problem

We need to calculate the heat lost by both water and ethanol as they cool from \(50^{\circ}C\) to \(10^{\circ}C\), and then compare these values.

Recall the Formula for Heat Transfer

The formula to calculate heat transfer is \( q = mc\Delta T \), where \( q \) is the heat lost or gained, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.

Calculate Heat Lost by Water

For water, \( m = 50.0 \text{ g} \), \( c = 4.18 \text{ J/g}\cdot\text{K} \), and \( \Delta T = 50^{\circ}C - 10^{\circ}C = 40^{\circ}C \). The heat lost by water, \( q_{\text{water}} \), is calculated as \[ q_{\text{water}} = 50.0 \times 4.18 \times 40 = 8360 \text{ J} \].

Calculate Heat Lost by Ethanol

For ethanol, \( m = 100.0 \text{ g} \), \( c = 2.46 \text{ J/g}\cdot\text{K} \), and \( \Delta T = 50^{\circ}C - 10^{\circ}C = 40^{\circ}C \). The heat lost by ethanol, \( q_{\text{ethanol}} \), is calculated as \[ q_{\text{ethanol}} = 100.0 \times 2.46 \times 40 = 9840 \text{ J} \].

Compare Heat Loss Values

Compare the heat lost values: \( q_{\text{water}} = 8360 \text{ J} \) and \( q_{\text{ethanol}} = 9840 \text{ J} \). Since \( q_{\text{ethanol}} > q_{\text{water}} \), ethanol gives up more heat.

Key concepts

Understanding Specific Heat Capacity

Specific heat capacity is a key concept in thermodynamics. It is the amount of heat required to change the temperature of one gram of a substance by one degree Celsius (or one Kelvin). The specific heat capacity is different for every material.

• Materials with high specific heat capacity require more energy to change their temperature.
• Water has a high specific heat capacity of 4.18 J/g·K, meaning it can absorb a lot of heat without a significant rise in temperature.
• Ethanol has a lower specific heat capacity of 2.46 J/g·K, meaning it heats up or cools down with less energy compared to water.

These differences influence how substances respond to heat and are crucial when comparing the heat given off by different materials.

Exploring Calorimetry

Calorimetry is the science of measuring the amount of heat released or absorbed during a physical or chemical process. It helps to understand energy transfer between substances.

• The process involves using a calorimeter to measure heat changes.
• In the problem, calorimetry concepts help us calculate the heat lost by water and ethanol as they cool from 50°C to 10°C.
• By applying the formula: \[q = mc\Delta T\]where \( q \) is heat lost, \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature, we assess energy transfer.

Knowing how calorimetry works allows us to compare and analyze different materials effectively.

Importance of Temperature Change in Heat Transfer

Temperature change, symbolized as \( \Delta T \), plays a vital role in heat transfer calculations. It signifies how much a substance's temperature will increase or decrease under specific conditions.

• The equation \( q = mc\Delta T \) shows the relationship between heat transferred, mass, specific heat capacity, and temperature change.
• In the context of the problem, both water and ethanol cool from 50°C to 10°C, causing a \( \Delta T \) of 40°C.
• This temperature change is crucial to calculating the heat given off because the larger the \( \Delta T \), the more heat is either absorbed or released.

Temperature changes drive the direction and amount of energy movement, thus are essential for predicting heat transfer between substances.