Questions: Prelaboratory Assignment 1. A 87.05 g sample of water at 94.6°C is mixed in a calorimeter with 46.50 g of water at 22.1°C. Calculate the final temperature of the water. 2. A piece of metal is heated by placing it in hot oil. (In lab it will be heated in hot water.) It is removed from the hot oil and dropped into a calorimeter of cold water. The water heats up due to the transfer of heat from the metal. a. After the metal has been moved to the calorimeter, the temperature of the calorimeter changes. Does it go up or down? b. How is the heat that causes the temperature of the calorimeter to change accounted for during our calorimetry calculations? 3. A sample of an ionic compound is dissolved in water and the water gets warmer. a. What substance is acting as the system? b. What substance is acting as the surroundings? c. Does heat flow from the system to the surroundings or from the surroundings to the system? d. Is dissolving this salt endothermic or exothermic? e. What is the sign of enthalpy for the dissolving process?

Solution Steps

To find the final temperature when two water samples are mixed, we use the principle of conservation of energy. The heat lost by the hot water will be equal to the heat gained by the cold water. The formula for heat transfer is:

\[ q = m \cdot c \cdot \Delta T \]

where \( q \) is the heat transferred, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. For water, \( c = 4.184 \, \text{J/g}^\circ\text{C} \).

Let \( T_f \) be the final temperature. The heat lost by the hot water is:

\[ q_{\text{hot}} = 87.05 \, \text{g} \cdot 4.184 \, \text{J/g}^\circ\text{C} \cdot (94.6^\circ\text{C} - T_f) \]

The heat gained by the cold water is:

\[ q_{\text{cold}} = 46.50 \, \text{g} \cdot 4.184 \, \text{J/g}^\circ\text{C} \cdot (T_f - 22.1^\circ\text{C}) \]

Setting \( q_{\text{hot}} = q_{\text{cold}} \) and solving for \( T_f \):

\[ 87.05 \cdot 4.184 \cdot (94.6 - T_f) = 46.50 \cdot 4.184 \cdot (T_f - 22.1) \]

Simplifying and solving for \( T_f \):

\[ 364.1 \cdot (94.6 - T_f) = 194.6 \cdot (T_f - 22.1) \]

\[ 34447.86 - 364.1T_f = 194.6T_f - 4295.66 \]

\[ 34447.86 + 4295.66 = 364.1T_f + 194.6T_f \]

\[ 38743.52 = 558.7T_f \]

\[ T_f = \frac{38743.52}{558.7} \approx 69.34^\circ\text{C} \]

When the metal is placed in the calorimeter, the temperature of the calorimeter goes up because the metal transfers heat to the water in the calorimeter.

The heat that causes the temperature change in the calorimeter is accounted for by measuring the temperature change of the water and using the specific heat capacity of water to calculate the heat absorbed. This heat is equal to the heat lost by the metal.

The system is the ionic compound being dissolved.

The surroundings are the water in which the ionic compound is dissolved.

Heat flows from the system (ionic compound) to the surroundings (water) because the water gets warmer.

Final Answer