Question

A laboratory technician drops a 0.0850-kg sample of unknown solid material, at 100.0\(^\circ\)C, into a calorimeter. The calorimeter can, initially at 19.0\(^\circ\)C, is made of 0.150 kg of copper and contains 0.200 kg of water. The final temperature of the calorimeter can and contents is 26.1\(^\circ\)C. Compute the specific heat of the sample.

Short answer

The specific heat of the sample is approximately 1.0 J/g°C.

Step-by-step solution

Understand the Problem

We are dealing with a calorimetry problem where an unknown solid is placed in a calorimeter, reaching thermal equilibrium. We are to find the specific heat of the solid sample using the given initial and final temperatures.

Write the Heat Exchange Equation

Since no heat is lost to the surroundings, we apply the conservation of heat principle: the heat lost by the sample is equal to the heat gained by the copper and water in the calorimeter.\[ m_s c_s (T_{s,i} - T_f) = m_c c_c (T_f - T_{c,i}) + m_w c_w (T_f - T_{w,i}) \]where:- \(m_s = 0.0850\,\text{kg}\)- \(T_{s,i} = 100.0\, ^\circ\text{C}\)- \(T_f = 26.1\, ^\circ\text{C}\)- \(m_c = 0.150\,\text{kg}\)- \(c_c = 0.385\, \text{J/g} ^\circ\text{C}\)- \(T_{c,i} = 19.0\, ^\circ\text{C}\)- \(m_w = 0.200\,\text{kg}\)- \(c_w = 4.186\, \text{J/g} ^\circ\text{C}\)- \(T_{w,i} = 19.0\, ^\circ\text{C}\)

Substitute Known Values

Substituting the known values into the equation:\[0.085 \cdot c_s \cdot (100.0 - 26.1) = 0.150 \cdot 0.385 \cdot (26.1 - 19.0) + 0.200 \cdot 4.186 \cdot (26.1 - 19.0)\]

Simplify and Solve for Specific Heat

Calculate each term separately:- Heat lost by the sample: \(0.085 \cdot c_s \cdot 73.9\)- Heat gained by the copper: \(0.150 \cdot 0.385 \cdot 7.1 = 0.410025\, \text{J}\)- Heat gained by the water: \(0.200 \cdot 4.186 \cdot 7.1 = 5.94352\, \text{J}\)Combine the heat gained by the copper and water:\[0.410025 + 5.94352 = 6.353545 \, \text{J}\]Set up the equation for the specific heat:\[0.085 \cdot c_s \cdot 73.9 = 6.353545\]Solve for \(c_s\):\[c_s = \frac{6.353545}{0.085 \times 73.9} \approx 1.0 \, \text{J/g} ^\circ\text{C}\]

Conclusion

The specific heat of the unknown solid is calculated to be approximately \(1.0 \, \text{J/g} ^\circ\text{C}\). This is a reasonable result compared to familiar substances.

Key concepts

Specific Heat Capacity

Specific heat capacity refers to the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius. Different materials require different amounts of heat to achieve the same temperature change. Metals, like copper, generally have low specific heats, meaning they warm up and cool down quickly. Water, on the other hand, has a high specific heat capacity, which allows it to absorb more heat before its temperature changes significantly.

Understanding the specific heat capacity is crucial when dealing with calorimetry problems—experiments designed to measure heat transfer. When a substance is introduced into a calorimeter, the specific heat capacity helps determine how much heat the substance will absorb or release to reach thermal equilibrium.

Heat Exchange Equation

The heat exchange equation is a mathematical representation of the heat transfer between substances involved in a calorimetry experiment. It is based on the principle that the total heat lost by hot objects must equal the total heat gained by cold objects, provided no heat is lost to the surroundings.

The equation used in the problem is expressed as:

• \( m_s c_s (T_{s,i} - T_f) = m_c c_c (T_f - T_{c,i}) + m_w c_w (T_f - T_{w,i}) \)

This equation is set up to calculate how much heat is gained or lost based on the specific heat capacities, masses, and temperature changes of the substances involved.

Thermal Equilibrium

Thermal equilibrium occurs when two or more substances reach the same temperature and no further heat transfer occurs between them. In the context of the calorimetry problem, reaching thermal equilibrium means that the heat lost by the hotter substance (unknown sample) is equal to the heat gained by the colder substances (calorimeter can and water).

Thermal equilibrium is important because it signifies the end point of the heat exchange process. Once thermal equilibrium is achieved, we know that all possible heat transfer has occurred and the temperatures will stabilize. In the exercise provided, thermal equilibrium was reached at 26.1\(^\circ\)C.

Conservation of Heat

Conservation of heat is a fundamental concept in thermodynamics, stating that energy cannot be created or destroyed, only transferred. In the context of calorimetry, this means that the total amount of heat exchanged in an isolated system remains constant.

The problem highlights the conservation of heat through the heat exchange equation, where the sum of the heat lost must equal the sum of the heat gained.

By assuming no heat is lost to the environment, we ensure that all heat accounted for is used in calculating the specific heat capacity, as seen in the example where total heat lost by the unknown solid is equal to total heat gained by the calorimeter's contents.