Problem 41 - An Introduction to Thermal Physics

Question

To measure the heat capacity of an object, all you usually have to do is put it in thermal contact with another object whose heat capacity you know. As an example, suppose that a chunk of metal is immersed in boiling water \(\left(100^{\circ} \mathrm{C}\right),\) then is quickly transferred into a Styrofoam cup containing \(250 \mathrm{g}\) of water at \(20^{\circ} \mathrm{C}\). After a minute or so, the temperature of the contents of the cup is \(24^{\circ} \mathrm{C} .\) Assume that during this time no significant energy is transferred between the contents of the cup and the surroundings. The heat capacity of the cup itself is negligible. (a) How much heat is lost by the water? (b) How much heat is gained by the metal? (c) What is the heat capacity of this chunk of metal? (d) If the mass of the chunk of metal is \(100 \mathrm{g}\), what is its specific heat capacity?

Step-by-Step Solution

Answer

(a) The water loses 418 J of heat. (b) The metal gains 418 J. (c) The heat capacity of the metal is 5.5 J/°C. (d) Its specific heat capacity is 0.055 J/g°C.

Step 1: Calculate Heat Lost by Water

The heat lost by the water can be calculated using the formula \( q = mc\Delta T \), where \( m \) is the mass of the water, \( c \) is the specific heat capacity of water (\( 4.18 \text{ J/g}^\circ\text{C} \)), and \( \Delta T \) is the change in temperature. Here, \( m = 250 \text{ g} \), \( T_i = 20^\circ\text{C} \), and \( T_f = 24^\circ\text{C} \). Thus, \( \Delta T = 24 - 20 \). Substituting in the values, we get:\[ q = 250 \times 4.18 \times (24 - 20) \]\[ q = 418 \text{ J} \]

Step 2: Heat Gained by the Metal

Since the energy is conserved and no significant energy is transferred to the surroundings, the heat gained by the metal must be equal to the heat lost by the water. Therefore, the heat gained by the metal is also \( 418 \text{ J} \).

Step 3: Calculate Heat Capacity of Metal

The heat capacity of the metal \( C_m \) is given by \( q = C_m \Delta T \), where \( \Delta T = 24^\circ\text{C} - 100^\circ\text{C} = -76^\circ\text{C} \). We set the heat gained by the metal equal to the lost heat, so \( 418 = C_m \times (-76) \). Solving for \( C_m \), we get:\[ C_m = \frac{418}{76} \]\[ C_m = 5.5 \text{ J/}^\circ\text{C} \]

Step 4: Calculate Specific Heat Capacity of Metal

The specific heat capacity \( c_m \) of the metal is given by \( c_m = \frac{C_m}{m} \), where \( m \) is the mass of the metal (\( 100 \text{ g} \)). Substituting the values, we have:\[ c_m = \frac{5.5}{100} \]\[ c_m = 0.055 \text{ J/g}^\circ\text{C} \]

Key Concepts

Heat Capacity

Heat capacity is a fundamental concept in thermal physics that describes the amount of heat required to change the temperature of an object by a certain amount.Heat capacity is an extensive property, which means it depends on the amount of substance present.The more of a substance you have, the greater its heat capacity.

To find an object's heat capacity, you often put it in thermal contact with another known object and observe the changes in temperature.This was precisely what was done in the exercise above: a piece of metal was immersed in water, and the temperature changes observed allowed for the determination of the metal's heat capacity.

• Heat capacity can be thought of as thermal inertia—it reflects how much an object resists changes in temperature.
• The formula for calculating heat capacity (\[ C = \frac{q}{\Delta T} \]) involves the amount of heat supplied or removed (\( q \)) and the resulting change in temperature (\( \Delta T \)).
• In the example, the heat capacity of the metal was found using the relationship between heat transfer and temperature change of the water and metal.

Specific Heat Capacity

Specific heat capacity narrows down the concept of heat capacity to a per-unit mass basis.It's a measure of how much heat energy it takes to raise 1 gram of a substance by 1 degree Celsius.This property is intensive, meaning it does not depend on the amount of substance, which makes it a very useful measure for identifying materials.

In the exercise, after determining the heat capacity of the metal, the specific heat capacity of the metal was found using its mass.This value allows for comparisons across different materials regardless of their amounts.For instance:

• The formula to calculate specific heat capacity is given by: \[ c = \frac{C}{m} \] where \( C \) is the heat capacity and \( m \) is the mass.
• This tells us that the chunk of metal, when considering its specific heat capacity, absorbs less heat per gram per degree than water.
• With a calculated specific heat capacity of the metal, one can predict how this metal will behave in other thermal scenarios.

Conservation of Energy

The principle of conservation of energy is pivotal in solving thermal physics problems such as this.It states that energy cannot be created or destroyed, only transformed from one form to another.In the context of the exercise, the energy lost by the water in the Styrofoam cup is gained by the metal, allowing us to solve for unknowns efficiently.

Assumptions, like no heat loss to the surroundings, simplify calculations by ensuring all the energy exchange happens between the known substances only.Here's what that means for the exercise:

• Energy lost by water (heat loss) must equal the energy gained by the metal (heat gain) when assuming no external energy transfers.
• Mathematically, this is expressed as: \[ q_{\text{lost, water}} = q_{\text{gained, metal}} \] where both must equate, leading to straightforward calculations of the heat interactions between the objects.
• This conservation principle ensures that we can trace energy flows and use them to solve thermal problems logically.