What is the energy change when the temperature of 10.0 grams of solid aluminum is decreased from 37.6 °C to 20.2 °C ?
The correct answer and explanation is:
Let’s solve the problem step-by-step:
Given:
- Mass of aluminum, m=10.0 gm = 10.0 \, \text{g}
- Initial temperature, Ti=37.6∘CT_i = 37.6^\circ C
- Final temperature, Tf=20.2∘CT_f = 20.2^\circ C
- Specific heat capacity of aluminum, c=0.897 J/g∘Cc = 0.897 \, \text{J/g}^\circ C (this is a known constant)
Goal:
Find the energy change qq when the temperature decreases from 37.6 °C to 20.2 °C.
Formula:
The energy change when a substance changes temperature (without a phase change) is: q=m×c×ΔTq = m \times c \times \Delta T
where
- qq is the heat energy absorbed or released (in joules),
- mm is mass (in grams),
- cc is specific heat capacity (in J/g°C),
- ΔT=Tf−Ti\Delta T = T_f – T_i is the change in temperature.
Step 1: Calculate ΔT\Delta T
ΔT=20.2−37.6=−17.4∘C\Delta T = 20.2 – 37.6 = -17.4^\circ C
The negative sign indicates a temperature decrease (cooling).
Step 2: Calculate energy change qq
q=10.0 g×0.897 J/g∘C×(−17.4∘C)=−156.2 Jq = 10.0 \, \text{g} \times 0.897 \, \text{J/g}^\circ C \times (-17.4^\circ C) = -156.2 \, \text{J}
Final Answer:
q=−156 J(rounded to 3 significant figures)q = -156 \, \text{J} \quad (\text{rounded to 3 significant figures})
The negative sign means energy is released (heat is lost by the aluminum as it cools down).
Explanation (300 words):
When the temperature of a substance changes, the amount of heat energy absorbed or released depends on three main factors: the mass of the substance, its specific heat capacity, and the temperature change it undergoes.
In this problem, we are dealing with solid aluminum, which has a known specific heat capacity of approximately 0.897 J/g°C. The specific heat capacity is the amount of energy required to raise the temperature of 1 gram of a substance by 1 degree Celsius.
Here, we want to find how much energy is lost when 10 grams of aluminum cools from 37.6°C down to 20.2°C. Since the temperature decreases, the aluminum releases heat energy.
The temperature change (ΔT\Delta T) is calculated by subtracting the initial temperature from the final temperature, resulting in -17.4°C. The negative sign indicates cooling or loss of heat.
Using the formula q=mcΔTq = m c \Delta T, we multiply the mass (10 g) by the specific heat capacity (0.897 J/g°C) and the temperature change (-17.4°C). This calculation yields a negative value (-156 J), confirming that the aluminum loses 156 joules of heat.
This energy change represents the heat transferred from the aluminum to its surroundings as it cools. Understanding such calculations is important in thermodynamics and engineering, where heat transfer and temperature control are crucial for designing systems, processes, and materials handling.
In summary, the aluminum releases about 156 joules of energy as it cools down by 17.4°C, which is a practical example of how substances exchange energy with their environment through temperature changes.