Problem 58 - Physics

Question

A thermos contains \(150 \mathrm{~cm}^{3}\) of coffee at \(85^{\circ} \mathrm{C}\). To cool the coffee, you drop two \(11-\mathrm{g}\) ice cubes into the thermos. The ice cubes are initially at \(0^{\circ} \mathrm{C}\) and melt completely. What is the final temperature of the coffee? Treat the coffee as if it were water.

Step-by-Step Solution

Answer

The final temperature of the coffee is approximately 37°C.

Step 1: Identify the thermodynamic system

Determine process type and knowns.

Step 2: Apply thermodynamic laws

Use PV=nRT, dU=Q-W, or entropy relations.

Step 3: Calculate

Substitute and solve.

Step 4: State the answer

The answer is: The final temperature of the coffee is approximately 37°C.

Key Concepts

Specific Heat Capacity

Specific heat capacity is a fundamental concept in thermodynamics. It tells us how much heat energy is needed to increase the temperature of a certain mass of a substance by one degree Celsius. The specific heat capacity of water, for example, is relatively high at 4.18 J/g°C. This means it takes 4.18 joules of energy to raise one gram of water by one degree Celsius.

When solving problems like the one in the exercise, knowing the specific heat capacity helps you calculate how much energy is stored in or released from a substance as its temperature changes.

• The formula for heat energy is:
  \( Q = m \, c \, \Delta T \)
• \(Q\) is the heat energy, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the temperature change.

In the exercise, the coffee's temperature change is what alters its thermal energy, and this change needs to be known to solve for the final temperature after adding ice.

Heat Transfer

Heat transfer is the movement of heat from a hotter object to a cooler one. In the case of the exercise, heat transfer occurs from the hot coffee to the cold ice cubes. The energy is transferred until thermal equilibrium is reached—that's when the coffee and the melted ice reach the same final temperature.

This process involves three different types of heat transfer: conduction, convection, and radiation. However, in this exercise, we are primarily concerned with conduction, as the heat is transferred directly between the coffee and the ice cubes when they come into contact.

• Because the coffee loses energy, its temperature decreases.
• The ice cubes absorb that energy, causing them to melt and eventually warm up to the final temperature of the coffee.

Understanding these points is crucial when calculating how much energy is transferred, which helps in determining the final temperature.

Phase Change

A phase change occurs when a substance changes from one state of matter to another, such as from solid to liquid. For the ice cubes in the exercise, this change from solid to liquid is key in solving the problem.

When ice melts, it undergoes a phase change and requires specific energy called the latent heat of fusion, which is the energy needed to change its state without increasing the temperature.

• The latent heat for ice is 334 J/g.
• This means each gram of ice needs 334 joules to melt while staying at 0°C.

In the exercise, both the ice melting and warming to the coffee temperature play roles in finding the energy transfer balance. Understanding a phase change helps you calculate the energy needed to melt the ice, which is subtracted from the coffee's energy, leading to the determination of the final temperature.