Math Problem Statement

0.50 mL of cooking oil is placed in the chamber of a constant-volume calorimeter filled with sufficient oxygen for the oil to be completely burned. The chamber is immersed in a water bath. The mass of the water bath is 2.15 kg. For the purpose of this analysis, the metal parts of the apparatus are modeled as equivalent to an additional 0.5 kg of water. The calorimeter is well-insulated, and initially the temperature throughout is 25°C. The oil is ignited by a spark, and when equilibrium is again attained, the calorie value of the cooking oil is -8.1 kcal/mL. Calculate temperature throughout, in °C, after equilibrium is once again attained.

Solution

We will use the principle of conservation of energy to solve this calorimetry problem. The heat released by the burning oil is absorbed by the water and the equivalent metal parts of the calorimeter.

Given data:

  • Volume of oil: 0.500.50 mL
  • Caloric value of oil: 8.1-8.1 kcal/mL (negative sign indicates heat released)
  • Mass of water: 2.152.15 kg
  • Mass of equivalent metal parts: 0.500.50 kg
  • Initial temperature: 25C25^\circ C
  • Specific heat capacity of water: 11 kcal/kg·°C
  • Total mass of water and equivalent metal parts:
    mtotal=2.15+0.50=2.65 kgm_{\text{total}} = 2.15 + 0.50 = 2.65 \text{ kg}

Step 1: Calculate the total heat released by burning oil

Qoil=(caloric value)×(volume of oil)Q_{\text{oil}} = (\text{caloric value}) \times (\text{volume of oil}) Qoil=(8.1kcal/mL)×(0.50mL)Q_{\text{oil}} = (-8.1 \, \text{kcal/mL}) \times (0.50 \, \text{mL}) Qoil=4.05kcalQ_{\text{oil}} = -4.05 \, \text{kcal}

Step 2: Use heat transfer equation

Since the heat is fully absorbed by the water and calorimeter, we apply:

Q=mcΔTQ = m c \Delta T

where
QQ is heat absorbed,
mm is the total mass,
cc is the specific heat of water,
ΔT\Delta T is the change in temperature.

Rearranging for ΔT\Delta T:

ΔT=Qmc\Delta T = \frac{Q}{m c}

Substituting values:

ΔT=4.052.65×1\Delta T = \frac{4.05}{2.65 \times 1}

ΔT=4.052.651.53C\Delta T = \frac{4.05}{2.65} \approx 1.53^\circ C

Step 3: Final Temperature

Tfinal=Tinitial+ΔTT_{\text{final}} = T_{\text{initial}} + \Delta T

Tfinal=25+1.53T_{\text{final}} = 25 + 1.53

Tfinal26.53CT_{\text{final}} \approx 26.53^\circ C

Final Answer:

The temperature after equilibrium is attained is approximately
26.53C\mathbf{26.53^\circ C}


Would you like me to break anything down further?

Here are five related questions you might find interesting:

  1. How would the final temperature change if the calorimeter had a different heat capacity?
  2. What happens if some heat is lost to the surroundings?
  3. How would the result change if we burned more oil?
  4. Can we use this method to determine the energy content of different fuels?
  5. How does a bomb calorimeter improve the accuracy of such experiments?

Tip: Always ensure the calorimeter is well-insulated to minimize heat loss and improve accuracy! 🔥

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Math Problem Analysis

Mathematical Concepts

Heat Transfer
Energy Conservation
Calorimetry
Specific Heat Capacity

Formulas

Q_oil = caloric value × volume of oil
Q = m c ΔT
ΔT = Q / (m c)
T_final = T_initial + ΔT

Theorems

Principle of Conservation of Energy

Suitable Grade Level

Grades 10-12