Question
A 50.0 g sample of Feat 100°C is added to 500.0 mL of water at 35°C in a perfectly insulated container. Which of the following statements is true?
(A) The temperature of the water will decrease.
(B) The temperature change of the Fe will have the same magnitude as the temperature change of the water.
(C) The heat energy lost by the Fe will be less than the heat energy gained by the water.
(D) The heat energy lost by the Fe will be equal to the heat energy gained by the water
▶️Answer/Explanation
.Ans:D
When the iron is added to the water, heat will transfer from the iron to the water until thermal equilibrium is reached.
\[ q = mc\Delta T \]
For iron (\( Fe \)):
\[ m_{Fe} = 50.0 \, \text{g} \]
\[ c_{Fe} = 0.449 \, \text{J/(g°C)} \] (specific heat capacity of iron)
\[ \Delta T_{Fe} = T_{\text{final, Fe}} – T_{\text{initial, Fe}} = T_{\text{final, Fe}} – 100°C \]
For water:
\[ m_{\text{water}} = 500.0 \, \text{g} \]
\[ c_{\text{water}} = 4.18 \, \text{J/(g°C)} \] (specific heat capacity of water)
\[ \Delta T_{\text{water}} = T_{\text{final, water}} – T_{\text{initial, water}} = T_{\text{final, water}} – 35°C \]
Now, let’s calculate the heat lost by the iron and gained by the water:
\[ q_{Fe} = m_{Fe} \times c_{Fe} \times \Delta T_{Fe} \]
\[ q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times \Delta T_{\text{water}} \]
Since energy is conserved, \( q_{Fe} = -q_{\text{water}} \).
Now, let’s analyze each statement:
(A) The temperature of the water will decrease.
True. Since heat transfers from the iron to the water, the temperature of the water will decrease.
(B) The temperature change of the Fe will have the same magnitude as the temperature change of the water.
False. The temperature change of the iron (\( \Delta T_{Fe} \)) will be different from the temperature change of the water (\( \Delta T_{\text{water}} \)) because the specific heat capacities are different.
(C) The heat energy lost by the Fe will be less than the heat energy gained by the water.
False. The heat energy lost by the iron will be equal in magnitude but opposite in sign to the heat energy gained by the water.
(D) The heat energy lost by the Fe will be equal to the heat energy gained by the water.
True. This statement is consistent with the principle of conservation of energy. The heat lost by the iron will be equal in magnitude but opposite in sign to the heat gained by the water.
Therefore, the correct answer is:(D).
Question
\(AgNO_{3}(aq)+NaCl(aq) \rightarrow AgCl(s)+NaNO_{3}\)(aq)
In an experiment a student mixes a 50.0 mL sample of 0.100 M \(AgNO_{3}\)(aq) with a 50.0 mL sample of 0.100 M NaCl(aq) at 20.0°C in a coffee-cup calorimeter. Which of the following is the enthalpy change of the precipitation reaction represented above if the final temperature of the mixture is 21.0°C? (Assume that the total mass of the mixture is 100. g and that the specific heat capacity of the mixture is 4.2 J/(g °C).)
(A) \(−84 kJ/mol_{rxn}\)
(B) \(−0.42 kJ/mol_{rxn}\)
(C) \(0.42 kJ/mol_{rxn}\)
(D) \(84 kJ/mol_{rxn}\)
▶️Answer/Explanation
Ans:C
Given:
Mass of the solution (\(m\)) = 100.0 g
Specific heat capacity of the solution (\(c\)) = 4.2 J/(g°C)
Change in temperature (\(\Delta T\)) = 1.0°C
Using the equation:
\[ q = m \times c \times \Delta T \]
\[ q = (100.0 \, \text{g}) \times (4.2 \, \text{J/(g°C)}) \times (1.0 \, \text{°C}) \]
\[ q = 420 \, \text{J} \]
To convert from joules to kilojoules:
\[ \Delta H = \frac{420 \, \text{J}}{1000 \, \text{J/kJ}} = 0.42 \, \text{kJ} \]
So, the enthalpy change of the precipitation reaction is indeed \(0.42 \, \text{kJ}\).
The correct answer is indeed:
(C) \(0.42 \, \text{kJ/mol}_{\text{rxn}}\)
Questions
When 5.0 g of \(NH^{4}ClO_{4(s)}\) is added to 100. mL of water in a calorimeter, the temperature of the solution formed decreases by 3.0\(^{\circ }C\). If 5.0 g of \(NH^{4}ClO_{4(s)}\) is added to 1000. mL of water in a calorimeter initially at 25.0\(^{\circ }C\), the final temperature of the solution will be approximately
(A) 22.0\(^{\circ }C\)
(B) 24.7\(^{\circ }C\)
(C) 25.3\(^{\circ }C\)
(D) 28.0\(^{\circ }C\)
▶️Answer/Explanation
Ans: B
Given:
When \(5.0 \, \text{g}\) of \(NH_4ClO_4(s)\) is added to \(100 \, \text{mL}\) of water, the temperature of the solution decreases by \(3.0^\circ C\).
The question asks about the final temperature when \(5.0 \, \text{g}\) of \(NH_4ClO_4(s)\) is added to \(1000 \, \text{mL}\) of water initially at \(25.0^\circ C\).
We can assume that the heat of solution for \(NH_4ClO_4(s)\) is an endothermic process, meaning that the dissolution absorbs heat from the surroundings, causing the temperature of the solution to decrease.
Since the temperature change is proportional to the amount of heat absorbed or released, and the heat of solution is a constant for a given compound, we can use the following relationship:
\[
\frac{\text{Temperature change for 100 mL of water}}{\text{Temperature change for 1000 mL of water}} = \frac{100 \, \text{mL}}{1000 \, \text{mL}} = \frac{1}{10}
\]
Therefore, the temperature change for \(1000 \, \text{mL}\) of water will be one-tenth of the temperature change for \(100 \, \text{mL}\) of water.
Given:
Temperature change for \(100 \, \text{mL}\) of water = \(-3.0^\circ C\)
Temperature change for \(1000 \, \text{mL}\) of water = \((-3.0^\circ C) \times \frac{1}{10} = -0.3^\circ C\)
Initial temperature of the solution = \(25.0^\circ C\)
Final temperature of the solution = Initial temperature + Temperature change
Final temperature of the solution = \(25.0^\circ C – 0.3^\circ C = 24.7^\circ C\)
Question
The graph above shows the changes in temperature recorded for the 2.00 L of H2O surrounding a constant-volume container in which a 1.00 g sample of benzoic acid was combusted. Assume that heat was not absorbed by the container or lost to the surroundings, that the density of H2O is 1.00g/mL, and that the specific heat capacity of H2O is about 4.2J/g⋅°C. Based on this information, estimate how much heat was released from the combustion of the benzoic acid sample.
A 0.013kJ
B 25kJ
C 180kJ
D 210kJ
▶️Answer/Explanation
Ans:B
The heat released from the combustion of benzoic acid was absorbed by the water and can be calculated by q=mcΔT . Estimating from the graph that the temperature change for the water was about 3.0°C and using 2000g as the mass of the water gives q=(2000g)×(4.2J/g⋅°C)×3.0°C=25,200 J, which is approximately equal to 25kJ.