**Q6.1.1**

Potential energy is usually described as the energy of position. Chemical potential energy is energy stored within the chemical bonds of a substance.

**Q6.1.2**

Answers will vary. The most common example in every day life is the burning of fossil fuels to generate electricity or to run a vehicle.

**Q6.1.3**

The temperature of the hot object decreases and the temperature of the cold object increases as heat is transferred from the hot object to the cold object. The change in temperature of each depends on the identity and properties of each substance.

**Q6.1.4**

The system is the specific portion of matter being observed in an experiment and is designated by the experimenter. The surroundings is everything that is not the system.

**Q6.1.5**

Endothermic processes result in the gain of heat to the system while exothermic processes are associated with the loss of heat from the system.

**Q6.1.6**

Reaction A is exothermic because heat is leaving the system making the test tube feel hot. Reaction B is endothermic because heat is being absorbed by the system making the test tube feel cold.

**Q6.1.7**

q is positive for endothermic processes and q is negative for exothermic processes.

**Q6.1.8**

Classify the following as endothermic or exothermic processes.

- Endothermic because heat is being added to the water to get it from the liquid state to the gas state.
- Endothermic because energy is consumed to evaporate the moisture on your skin which lowers your temperature.
- Exothermic because burning (also known as combustion) releases heat.
- Exothermic because energy is exiting the system in order to go from liquid to solid. Another way to look at it is to consider the opposite process of melting. Energy is consumed (endothermic) to melt ice (solid to liquid) so the opposite process (liquid to solid) must be exothermic.

**Q6.1.9**

Convert each value to the indicated units.

- \(150\; kcal\left ( \frac{1\;Cal}{1\;kcal} \right )=150\;Cal\)
- \(355\;J\left ( \frac{1\;cal}{4.184\;J} \right )=84.8\;cal\)
- \(200.\;Cal\left ( \frac{1000\;cal}{1\;Cal} \right )\left ( \frac{4.184\;J}{1\;cal} \right )=8.37\times 10^{5}\;J\)
- \(225\;kcal\left ( \frac{1000\;cal}{1\;kcal} \right )=2.25\times 10^{5}\;cal\)
- \(3450.\;cal\left ( \frac{1 kcal\;kJ}{1000\;cal} \right )=3.450\;kcal\)
- \(450.\;Cal\left ( \frac{1000\;cal}{1\;Cal} \right )\left ( \frac{4.184\;J}{1\;cal} \right )\left ( \frac{1\;kJ}{1000\;J} \right )=1.88\times 10^{3}\;kJ\) or \(450.\;Cal\left ( \frac{4.184\;kJ}{1\;Cal} \right )=1.88\times 10^{3}\;kJ\)
- \(175\;kJ\left ( \frac{1000\;J}{1\;kJ} \right )\left ( \frac{1\;cal}{4.184\;J} \right )=4.18\times 10^{4}\;cal\)

**Q6.1.10**

Iron has a specific heat capacity of 0.449 \(\text{J/g} \cdot ^\text{o} \text{C}\) which means it takes 0.449 J of energy to raise 1 gram of iron by 1\(^\text{o} \text{C}\). Aluminum has a specific heat capacity of 0.897 \( \text{J/g} \cdot ^\text{o} \text{C} \) which means it takes 0.897 J of energy to raise 1 gram of aluminum by 1\(^\text{o} \text{C}\). When equal amounts of heat are applied, the temperature of the iron will increase more because it takes less energy (heat) to raise its temperature so iron will be at a higher temperature since they both start at 25\(^\text{o} \text{C}\).

**Q6.1.11**

Both samples are the same mass so a comparison of the specific heat must be compared. Copper has a specific heat of 0.385 \(\text{J/g} \cdot ^\text{o} \text{C}\) which means it takes 0.385 J of energy to raise 1 gram of copper by 1^\text{o} \text{C}\). Lead has a specific heat of 0.129 \(\text{J/g} \cdot ^\text{o} \text{C}\) which means it takes 0.129 J of energy to raise 1 gram of copper by 1^\text{o} \text{C}\). More energy is needed to raise the temperature of copper so more heat will be needed to increase the temperature of copper by 10 ^\text{o} \text{C}\).

**Q6.1.12**

\(\begin{array}{c}

q=m\cdot C_{p}\cdot \Delta T\\

q=50.0\;g\cdot 0.233\; \frac{J}{g\cdot^\text{o} \text{C} }\cdot \left (50^\text{o} \text{C}-30^\text{o} \text{C} \right )\\

q=50.0\;g\cdot 0.233\;\frac{J}{g\cdot^\text{o} \text{C} }\cdot \left (20^\text{o} \text{C} \right )\\

q=233\; J

\end{array}\)

**Q6.1.13**

\(\begin{array}{c}

q=m\cdot C_{p}\cdot \Delta T\\

125\;J=20.0\;g\cdot 0.129\;\frac{J}{g\cdot^\text{o} \text{C} }\cdot \left (T_{f}-15^\text{o} \text{C} \right )\\

48.4^\text{o} \text{C}=T_{f}-15^\text{o} \text{C}\\

T_{f}=63^\text{o} \text{C}

\end{array}\)

**Q6.1.14**

\(\begin{array}{c}

q=m\cdot C_{p}\cdot \Delta T\\

q=13.7\;g\cdot 0.897\;\frac{J}{g\cdot^\text{o} \text{C} }\cdot \left (61.9^\text{o} \text{C}-25.2^\text{o} \text{C} \right )\\

q=13.7\;g\cdot 0.897\;\frac{J}{g\cdot^\text{o} \text{C} }\cdot \left (36.7^\text{o} \text{C} \right )\\

q=451\; J

\end{array}\)

**Q6.1.15**

\(\begin{array}{c}

q=m\cdot C_{p}\cdot \Delta T\\

2250\;J=274\;g\cdot C_{p}\ \cdot 8.11^\text{o} \text{C}\\

C_{p}=1.01 \frac{J}{g\cdot^\text{o} \text{C} }

\end{array}\)

**Q6.1.16**

\(\begin{array}{c}

q=m\cdot C_{p}\cdot \Delta T\\

98.3\;J=12.28\;g\cdot C_{p}\ \cdot 5.42^\text{o} \text{C}\\

C_{p}=1.48 \frac{J}{g\cdot^\text{o} \text{C} }

\end{array}\)

**Q6.1.17**

\(\begin{array}{c}

q=m\cdot C_{p}\cdot \Delta T\\

-3.12\;kJ=m\cdot 2.44\frac{J}{g\cdot^\text{o} \text{C} }\left (12.3^\text{o} \text{C}-47.9^\text{o} \text{C} \right )\\

-3.12\times10^{3}\;J=m\cdot 2.44\frac{J}{g\cdot^\text{o} \text{C} }\left (-35.6^\text{o} \text{C} \right )\\

m=35.9\;g

\end{array}\)