Problem 46 - College Physics

Question

A laboratory technician drops an \(85.0 \mathrm{~g}\) solid sample of unknown material at a temperature of \(100.0^{\circ} \mathrm{C}\) into a calorimeter. The calorimeter can is made of \(0.150 \mathrm{~kg}\) of copper and contains \(0.200 \mathrm{~kg}\) of water, and both the can and water are initially at \(19.0^{\circ} \mathrm{C}\). The final temperature of the system is measured to be \(26.1^{\circ} \mathrm{C}\). Compute the specific heat of the sample. (Assume no heat loss to the surroundings.)

Step-by-Step Solution

Answer

The specific heat of the sample is approximately 1004 J/kg°C.

Step 1: Determine Heat Gained by Water

First, we calculate the amount of heat gained by the water in the calorimeter. The formula to calculate heat is given by \[ q = m \times c \times \Delta T \]where:- \( q \) is the heat gained,- \( m \) is the mass of the substance,- \( c \) is the specific heat capacity of the substance, and- \( \Delta T \) is the change in temperature.For water:- \( m = 0.200 \text{ kg} \)- \( c = 4186 \text{ J/kg°C} \) (specific heat of water),- \( \Delta T = 26.1^{\circ} \text{C} - 19.0^{\circ} \text{C} = 7.1^{\circ} \text{C} \).Plugging these values into the formula:\[ q_{\text{water}} = 0.200 \times 4186 \times 7.1 \approx 5944.12 \text{ J} \].

Step 2: Determine Heat Gained by Copper Can

Now, calculate the heat gained by the copper calorimeter can. Use the same heat equation \[ q = m \times c \times \Delta T \].For copper:- \( m = 0.150 \text{ kg} \)- \( c = 385 \text{ J/kg°C} \) (specific heat of copper),- \( \Delta T = 26.1^{\circ} \text{C} - 19.0^{\circ} \text{C} = 7.1^{\circ} \text{C} \).Calculate:\[ q_{\text{copper}} = 0.150 \times 385 \times 7.1 \approx 409.275 \text{ J} \].

Step 3: Apply Conservation of Energy

The conservation of energy principle states that the heat lost by the sample will be equal to the heat gained by the water and the copper can.Since there is no heat lost to the surroundings,\[ q_{\text{sample}} = -(q_{\text{water}} + q_{\text{copper}}) \].So the heat lost by the sample is\[ q_{\text{sample}} = - (5944.12 + 409.275) = -6353.395 \text{ J} \].

Step 4: Calculate Specific Heat of the Sample

To find the specific heat of the sample, use the formula \[ q = m \times c \times \Delta T \].Rearrange to solve for \( c \):\[ c = \frac{q}{m \times \Delta T} \].Here, for the sample:- \( q = -6353.395 \text{ J} \) (negative because the sample is losing heat),- \( m = 0.085 \text{ kg} \),- \( \Delta T = 100.0^{\circ} \text{C} - 26.1^{\circ} \text{C} = 73.9^{\circ} \text{C} \).Plug these values into the formula:\[ c = \frac{-6353.395}{0.085 \times 73.9} \approx 1004.21 \text{ J/kg°C} \].

Key Concepts

Understanding Specific Heat Capacity

Specific heat capacity is a crucial concept when dealing with calorimetry problems. It defines how much heat energy is required to change the temperature of a substance by one degree Celsius per unit mass.

This property depends on the material’s nature, which means that different materials absorb heat differently. For instance, water has a high specific heat capacity of 4186 J/kg°C, meaning it can absorb a lot of heat without undergoing a significant temperature change. Copper, on the other hand, with a lower specific heat capacity of 385 J/kg°C, heats up more easily.

• Specific heat helps determine how substances in a calorimeter will respond to heat exchange.
• Different materials react uniquely due to varying molecular structures and bonding.

Understanding specific heat capacity gives insights into why substances change temperature at different rates when exposed to heat and allows us to compute the energy exchanges during thermal processes.

Conservation of Energy in Calorimetry

The conservation of energy is a fundamental principle in physics, stating that energy within a closed system must remain constant, meaning energy lost by one part of the system must be gained by another.

In calorimetry, this principle ensures that all the heat lost by a hot object is gained by cooler objects, without any loss to the surroundings. In the provided exercise, the unknown sample loses heat, which is then gained by the water and copper calorimeter can.

• Heat lost = Heat gained: The sum of energy produced equals the sum of energy absorbed.
• Ensures precise calculations of unknown variables, such as the specific heat of an unknown sample.

This principle is used to set up equations that allow us to solve for unknowns, such as specific heat capacity, using the known values of other components involved in the process.

Mechanics of Heat Transfer

Heat transfer is the process of energy exchange from a higher temperature object to a cooler one, essential in calorimetry.
There are three primary modes of heat transfer: conduction, convection, and radiation. In the context of the exercise, conduction is usually what's considered, as it involves the direct transfer of heat through contact, without moving material.

• In the calorimeter, heat is transferred primarily through conduction, from the unknown sample to the water and copper can.
• The efficiency of heat transfer influences the accuracy of calorimetry measurements.

Understanding heat transfer is vital for calculating how much heat a substance can give or receive, which directly ties into the calculations needed to find specific heat capacity.

Achieving Thermal Equilibrium

Thermal equilibrium is reached when all parts of a system stabilize at a uniform temperature and no more heat transfer occurs.
In our exercise, this state is achieved when the drop in temperature of the heated sample perfectly balances with the rise in temperature of the water and the copper can.

• Temperature equalization means all components stop changing temperature.
• This end state is crucial to accurately measure heat transfer and validate calculations when testing for specific heat capacities.

Understanding thermal equilibrium helps us know when to take final measurements in a calorimetric experiment, ensuring reliable and precise results.