Problem 11
Question
The initial temperature of a 344 -g sample of iron is \(18.2^{\circ} \mathrm{C} .\) If the sample absorbs \(2.25 \mathrm{kJ}\) of heat, what is its final temperature?
Step-by-Step Solution
Verified Answer
The final temperature of the iron is approximately 32.77°C.
1Step 1: Understand the problem
We need to determine the final temperature of a 344-g sample of iron initially at 18.2°C after it absorbs 2.25 kJ of heat.
2Step 2: Identify the formula
We use the formula for calculating heat change, which is \( q = mc\Delta T \), where \( q \) is the heat added, \( m \) is the mass of the substance, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.
3Step 3: Gather known values
- \( m = 344 \) g (iron's mass) - \( q = 2.25 \) kJ (heat absorbed, convert this to J: \( 2.25 \times 1000 = 2250 \) J) - \( c = 0.449 \) J/g°C (specific heat capacity of iron) - Initial temperature \( T_i = 18.2^{\circ} \mathrm{C} \) (Given)- Final temperature \( T_f \) (Needs to be calculated)
4Step 4: Use the formula to find \( \Delta T \)
Rearrange the formula to solve for \( \Delta T \): \( \Delta T = \frac{q}{mc} \) Substitute the known values: \( \Delta T = \frac{2250}{344 \times 0.449} \) \( \Delta T = \frac{2250}{154.456} \) \( \Delta T \approx 14.57^{\circ} \mathrm{C} \) (rounded to two decimal places).
5Step 5: Calculate the final temperature \( T_f \)
To find the final temperature, use the formula: \( T_f = T_i + \Delta T \) Substitute the known values: \( T_f = 18.2 + 14.57 \) \( T_f \approx 32.77^{\circ} \mathrm{C} \) (rounded to two decimal places).
Key Concepts
Heat TransferTemperature ChangeCalorimetry
Heat Transfer
Heat transfer is a fundamental concept in physics that describes the movement of energy from one object to another because of a temperature difference. When heat flows into or out of an object, it causes a change in the energy contained within that object. In our exercise, the 344-g sample of iron absorbs 2.25 kJ of heat, illustrating the process of heat transfer.
There are three main modes of heat transfer:
There are three main modes of heat transfer:
- Conduction: Heat transfer through direct contact between materials, like a metal spoon getting hot when placed in a pot of boiling water.
- Convection: Heat transfer through fluid motion, such as warm air rising and cool air descending in a room.
- Radiation: Heat transfer via electromagnetic waves, like the warmth from sunlight.
Temperature Change
The change in temperature (\(\Delta T\)) occurs when an object either gains or loses heat energy. Specifically, in the context of our exercise, the temperature of the iron increases as it absorbs heat. The relationship between heat transfer and temperature change can be calculated using the formula: \( q = mc\Delta T \),where:
- \(q\) is the heat energy added or removed,
- \(m\) is the mass,
- \(c\) is the specific heat capacity,
- \(\Delta T\) is the temperature change.
Calorimetry
Calorimetry is the science of measuring the amount of heat involved in a chemical or physical process. It can be used to measure changes in temperature, phase changes, or chemical reactions. In this exercise, calorimetry allows us to calculate the final temperature of the iron sample after absorbing heat.In essence, calorimetry involves the use of a calorimeter to compute the heat exchanged during physical changes like melting or heating. By knowing the specific heat capacity, mass, and amount of heat added, we can find out how much the temperature will change in the substance.
Components in calorimetry:
Components in calorimetry:
- **Calorimeter:** A device used to measure temperature changes.
- **Temperature measurements:** Initial and final temperature readings are key to finding heat change.
- **Specific Heat Capacity (\(c\):**A crucial value that indicates how much heat is needed to change the temperature of a unit mass by 1°C.
Other exercises in this chapter
Problem 8
The specific heat capacity of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is \(1.74 \mathrm{J} / \mathrm{g} \cdot \mathrm{K} . \text { What is its mo
View solution Problem 9
The specific heat capacity of copper is \(0.385 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\). What quantity of heat is required to heat 168 g of copper from \(-12
View solution Problem 12
After absorbing \(1.850 \mathrm{kJ}\) of heat, the temperature of a \(0.500-\mathrm{kg}\) block of copper is \(37^{\circ} \mathrm{C} .\) What was its initial te
View solution Problem 14
A 182 -g sample of gold at some temperature is added to 22.1 g of water. The initial water temperature is \(25.0^{\circ} \mathrm{C}\) and the final temperature
View solution