Question

A 110.0-g sample of a gray-colored, unknown, pure metal was heated to \(92.0^{\circ} \mathrm{C}\) and put into a coffee-cup calorimeter containing \(75.0 \mathrm{~g}\) of water at \(21.0^{\circ} \mathrm{C}\). When the heated metal was put into the water, the temperature of the water rose to a final temperature of \(24.2^{\circ} \mathrm{C}\). The specific heat of water is \(4.184 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\). (a) What is the specific heat of the metal? (b) Is it possible that the metal is either iron or lead? Explain.

Short answer

(a) 0.135 J/g°C. (b) The metal is likely lead.

Step-by-step solution

- Determine the heat gained by water

Use the formula for heat gained by water: \ \[ q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times \triangle T_{\text{water}} \] \where \ \( m_{\text{water}} \) is the mass of water, \ \( c_{\text{water}} \) is the specific heat of water, and \ \( \triangle T_{\text{water}} \) is the change in temperature of water: \ \( \triangle T_{\text{water}} = 24.2^{\text{\circ} \text{C}} - 21.0^{\text{\circ} \text{C}} = 3.2^{\text{\circ} \text{C}} \). Substitute the values: \ \( q_{\text{water}} = 75.0 \text{ g} \times 4.184 \text{ J/g}^{\text{\circ} \text{C}} \times 3.2^{\text{\circ} \text{C}} = 1003.2 \text{ J} \).

- Determine the heat lost by the metal

The heat lost by the metal is equal to the heat gained by water since no heat is lost to the surroundings. Therefore, \ \( q_{\text{metal}} = -q_{\text{water}} = -1003.2 \text{ J} \).

- Determine the specific heat of the metal

Use the formula for heat lost by the metal: \ \[ q_{\text{metal}} = m_{\text{metal}} \times c_{\text{metal}} \times \triangle T_{\text{metal}} \] \where \ \( m_{\text{metal}} \) is the mass of metal (110.0 g), \ \( \triangle T_{\text{metal}} \) is the change in temperature of the metal (\( 92.0^{\text{\circ} \text{C}} - 24.2^{\text{\circ} \text{C}} = 67.8^{\text{\circ} \text{C}} \)), and \ \( c_{\text{metal}} \) is the specific heat of the metal. Substitute the known values: \ \( -1003.2 \text{ J} = 110.0 \text{ g} \times c_{\text{metal}} \times 67.8^{\text{\circ} \text{C}} \). Solve for the specific heat of the metal: \ \[ c_{\text{metal}} = \frac{-1003.2 \text{ J}}{110.0 \text{ g} \times 67.8^{\text{\circ} \text{C}}} = 0.135 \text{ J/g}^{\text{\circ} \text{C}} \].

- Determine if the metal is iron or lead

Compare the calculated specific heat of the metal (0.135 J/g°C) with the known specific heats of iron (0.449 J/g°C) and lead (0.128 J/g°C). Since the calculated specific heat is much closer to the specific heat of lead, it is likely that the unknown metal is lead.

Key concepts

heat transfer

Heat transfer occurs when thermal energy moves from one substance to another. This typically happens due to a temperature difference. Heat always flows from the hotter substance to the colder one until thermal equilibrium is reached. In calorimetry, we often look at the heat transferred so we can calculate other properties, like specific heat capacity. For example, in the exercise given, the heat transfer caused the water's temperature to rise when the hot metal was placed in it. This change allows us to understand how much heat was transferred from the metal to the water.

calorimetry

Calorimetry is a method used to measure the amount of heat transferred to or from a substance. It involves using a calorimeter, a device that helps us measure heat changes during physical and chemical processes. In this exercise, a coffee-cup calorimeter was used. This type of calorimeter is simple and consists of two nested Styrofoam cups, with a lid and a thermometer. When the sample metal was placed in the water inside the calorimeter, the temperature change in the water was used to calculate the heat absorbed by the water. From this, we deduce the heat lost by the metal, helping us find its specific heat capacity.

specific heat capacity

Specific heat capacity is an important concept in thermal physics. It indicates the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius. The formula used to calculate specific heat capacity is given by \[ q = m \times c \times \triangle T \] where \( q \) represents the heat transferred, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \triangle T \) is the change in temperature. For example, in the provided exercise, we calculated the specific heat of the metal by rearranging this formula and solving it using the given values. The specific heat capacity of the unknown metal was found to be 0.135 J/g°C.

metals identification

Metals identification involves determining which metal we are dealing with based on its physical properties, such as specific heat capacity. In our exercise, we calculated the specific heat capacity of an unknown metal and compared this with known values for iron and lead. The value calculated for the unknown metal was closer to the specific heat capacity of lead, which allowed us to identify the metal as likely being lead. This process is crucial in various applications, where knowing the exact type of metal can impact the usage, handling, and processing of the material.