Calculating Heat Transfer in Edible Oil

This problem asks us to calculate the amount of heat energy required to raise the temperature of a specific mass of edible oil. This involves the concept of specific heat capacity, which is a physical property of a substance that tells us how much energy is needed to raise the temperature of 1 kg of that substance by 1 Kelvin (or 1 degree Celsius).

The formula used to calculate the heat transfer (\(Q\)) is:

\(Q = mc\Delta T\)

Where:

• \(Q\) is the heat energy transferred (in Joules).
• \(m\) is the mass of the substance (in kilograms).
• \(c\) is the specific heat capacity of the substance (in J kg^-1 K^-1).
• \(\Delta T\) is the change in temperature (in Kelvin or degrees Celsius).

Identifying Given Values for Heat Calculation

From the question, we are given the following information:

• Mass of edible oil (\(m\)) = 100 g
• Specific heat capacity of edible oil (\(c\)) = 1965 J kg^-1 K^-1
• Change in temperature (\(\Delta T\)) = 20 °C

Converting Units for Accurate Heat Transfer Calculation

Before plugging the values into the formula, we need to ensure all units are consistent. The mass is given in grams, but the specific heat capacity is in J kg^-1 K^-1. We must convert the mass from grams to kilograms.

• 1 kg = 1000 g
• Mass (\(m\)) = 100 g = \(100 \div 1000\) kg = 0.1 kg

The temperature change is given in degrees Celsius. A temperature change of 20°C is equivalent to a temperature change of 20 K, so \(\Delta T = 20\) K.

Performing the Heat Transfer Calculation

Now we can substitute the values into the heat transfer formula \(Q = mc\Delta T\):

\(Q = (0.1 \text{ kg}) \times (1965 \text{ J kg}^{-1} \text{ K}^{-1}) \times (20 \text{ K})\)

\(Q = 0.1 \times 1965 \times 20 \text{ J}\)

\(Q = 196.5 \times 20 \text{ J}\)

\(Q = 3930 \text{ J}\)

The question asks for the answer in kilojoules (kJ). To convert Joules to kilojoules, we divide by 1000.

\(Q = 3930 \div 1000 \text{ kJ}\)

\(Q = 3.93 \text{ kJ}\)

Comparing Calculated Heat with Options

The calculated heat required is 3.93 kJ. Let's compare this with the given options:

Our calculated value of 3.93 kJ matches Option 4.

Summary of Heat Transfer Solution

To find the heat needed to raise the temperature of the edible oil, we used the heat transfer formula \(Q=mc\Delta T\), ensuring mass was in kilograms and temperature change was compatible with the specific heat units.

• Mass (m) = 100 g = 0.1 kg
• Specific heat (c) = 1965 J/kg·K
• Temperature change (\(\Delta T\)) = 20 °C = 20 K
• \(Q = (0.1 \text{ kg}) \times (1965 \text{ J/kg·K}) \times (20 \text{ K})\)
• \(Q = 3930 \text{ J}\)
• \(Q = 3.93 \text{ kJ}\)

Therefore, 3.93 kJ of heat should be transferred to the 100 g of edible oil to raise its temperature by 20°C.

Revision Table: Key Concepts for Heat Transfer

Additional Information: Specific Heat and Thermal Energy

Specific heat capacity is an intensive property, meaning it does not depend on the amount of substance. It's a measure of how resistant a substance is to changing its temperature when heat is added or removed. Materials with a high specific heat capacity require more energy to change their temperature compared to materials with a low specific heat capacity. Water, for instance, has a very high specific heat capacity (around 4186 J/kg·K), which is why it is used in cooling systems and takes a long time to heat up or cool down. Oils generally have lower specific heat capacities than water.

The calculation \(Q = mc\Delta T\) assumes that no phase change (like melting or boiling) occurs during the temperature change. It also assumes that the specific heat capacity remains constant over the temperature range considered, which is a good approximation for small temperature changes. This calculation is fundamental in understanding how thermal energy is transferred and stored in materials.