Question

A 182 -g sample of gold at some temperature was added to 22.1 g of water. The initial water temperature was \(25.0^{\circ} \mathrm{C},\) and the final temperature was \(27.5^{\circ} \mathrm{C} .\) If the specific heat capacity of gold is \(0.128 \mathrm{J} / \mathrm{g} \cdot \mathrm{K},\) what was the initial temperature of the gold sample?

Short answer

The initial temperature of the gold was 30.0°C.

Step-by-step solution

Understand the Formula

We are looking to find the initial temperature of gold. For this, we use the formula for heat transfer: \[ q = m \cdot c \cdot \Delta T \] where \( q \) is the heat energy transferred, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. Heat lost by gold must equal heat gained by water.

Calculate Heat Gained by Water

First, calculate the heat gained by the water using its mass and specific heat capacity (which is known to be \( 4.18 \mathrm{J/g\cdot K} \)). Since the temperature change \( \Delta T \) is the final temperature minus the initial temperature, for water:\[ \Delta T_{\text{water}} = 27.5^{\circ} C - 25.0^{\circ} C = 2.5^{\circ} C \] Thus, heat gained by water is:\[ q_{\text{water}} = 22.1 \cdot 4.18 \cdot 2.5 \]

Solve for Heat Gained by Water

Perform the calculations for the heat gained by the water:\[ q_{\text{water}} = 22.1 \cdot 4.18 \cdot 2.5 = 230.35 \text{ J} \] This is the amount of heat the water gained.

Set Up Heat Equation for Gold

Using the fact that the heat lost by the gold equals the heat gained by the water, we set:\[ q_{\text{gold}} = q_{\text{water}} = 230.35 \text{ J} \] Now, use the equation for \( q \) in terms of gold:\[ 230.35 = 182 \cdot 0.128 \cdot (T_{\text{gold initial}} - 27.5) \] "},{

Key concepts

Specific Heat Capacity

Specific heat capacity is a property of a material that indicates how much heat energy is needed to raise the temperature of one gram of the substance by one degree Celsius (or one Kelvin). It is expressed in units of Joules per gram per degree Celsius (J/g °C) or per Kelvin. This property helps us understand how different substances absorb or release heat differently.

In the context of our problem, the specific heat capacity of gold is given as 0.128 J/g·K. This means gold requires 0.128 Joules to increase its temperature by 1 K for each gram. Comparing this with water, which has a specific heat capacity of 4.18 J/g·K, gold heats up and cools down quicker than water because it requires less energy per unit mass to change its temperature.

Understanding specific heat capacity is crucial in predicting how substances will react in thermal exchanges. Substances with high specific heat capacity, like water, are efficient at storing heat, whereas those with low specific heat capacities, like metals, do not store heat as efficiently. This is why water is often used as a coolant.

Temperature Change

Temperature change, denoted as \( \Delta T \), represents the difference between the initial and final temperatures of a substance in a heat transfer scenario. It indicates how much the temperature has increased or decreased due to the addition or removal of heat.

In our problem, the temperature change of the water \( \Delta T_{\text{water}} = 2.5^{\circ} C \) is calculated by subtracting the initial temperature from the final temperature. This temperature change is used in the formula for heat transfer \( q = m \cdot c \cdot \Delta T \) to calculate the amount of heat gained or lost by a substance.

For gold, we needed to determine its initial temperature. Knowing the heat exchanges between gold and water occur until thermal equilibrium is reached, we use the equation \( q_{\text{gold}} = q_{\text{water}} \) to solve for the initial temperature of gold. The process of calculating specific temperature changes involves understanding energy conservation during thermal exchanges.

Calorimetry

Calorimetry is the science of measuring the heat of chemical reactions or physical changes. It involves observing the temperature change when a body absorbs or releases heat, and it hinges on the principles of energy conservation.

In our exercise, we applied calorimetry to calculate the initial temperature of the gold sample by measuring the heat transfer to and from water. By knowing the mass of the substances involved, their specific heat capacities, and the change in temperature (\( \Delta T \)), we set up equations based on the principle that the heat lost by gold equals the heat gained by water.

Calorimetry is a powerful tool in chemistry and physics for studying thermal properties and energy changes, as it uses relatively simple experimental setups to provide insights into exothermic and endothermic processes. This method helps predict how different materials will react to temperature variations, which is vital for applications like designing thermal systems or understanding climatic phenomena.