Problem 77 - Chemistry Matter and Change

Question

Food Preparation When 10.2 g of canola oil at 25.0°C is placed in a wok, 3.34 kJ of heat is required to heat it to a temperature of 196.4°C. What is the specific heat of the canola oil?

Step-by-Step Solution

Answer

The specific heat of the canola oil is approximately 1.92 J/g°C.

Step 1: Identify the given values

We are given the following values: - mass of canola oil (m) = 10.2 g - initial temperature (T1) = 25.0°C - final temperature (T2) = 196.4°C - heat required to heat the oil (q) = 3.34 kJ First, we need to convert the heat unit from kJ to J (joules) since specific heat is typically measured in J/g°C. To convert kJ to J, we'll multiply by 1000 since 1 kJ = 1000 J. q = 3.34 kJ * 1000 = 3340 J Now, our known values are: - m = 10.2 g - T1 = 25.0°C - T2 = 196.4°C - q = 3340 J

Step 2: Calculate the temperature change

Find the change in temperature (ΔT) by subtracting the initial temperature (T1) from the final temperature (T2). ΔT = T2 - T1 ΔT = 196.4°C - 25.0°C ΔT = 171.4°C

Step 3: Use the heat transfer equation to find the specific heat capacity

Now we have all the needed values to find the specific heat capacity of canola oil. Rearrange the heat transfer equation to solve for the specific heat capacity (c): c = q / (m * ΔT) Substitute the values: c = 3340 J / (10.2 g * 171.4°C) c ≈ 1.92 J/g°C The specific heat of the canola oil is approximately 1.92 J/g°C.

Key Concepts

Heat Transfer Equation

In physics and chemistry, the heat transfer equation is a crucial tool. It's used to determine the energy required to change the temperature of a substance. This equation is expressed as:\[ q = m \cdot c \cdot \Delta T \]Here, \( q \) is the total heat energy added or removed, \( m \) stands for mass, \( c \) represents specific heat capacity, and \( \Delta T \) is the change in temperature. This equation essentially helps in understanding how much energy is needed to heat a particular mass of a substance by a certain temperature.By rearranging this formula, we can find any of its components as long as the other values are known. It's particularly useful in experiments and real-world situations like cooking or material testing.

Temperature Change

Temperature change, denoted as \( \Delta T \), is an important factor when calculating heat transfer. It tells us how much we need to increase or decrease a substance's temperature. To find this value, subtract the initial temperature from the final temperature:\[ \Delta T = T_2 - T_1 \]In our case, the initial temperature \( (T_1) \) was 25.0°C and the final temperature \( (T_2) \) was 196.4°C, resulting in a change of 171.4°C.
This significant change indicates how much energy was required to achieve the final desired state of the canola oil. It's crucial always to monitor the temperature accurately to ensure precise calculations in thermal exercises.

Unit Conversion

Unit conversions are an integral part of scientific calculations. They help in ensuring that all variables in an equation are compatible with each other. In thermal calculations, heat is often measured in joules (J) rather than kilojoules (kJ). Because the specific heat is usually given in J/g°C, conversion is necessary when the heat quantity is provided in kJ. To convert kilojoules to joules, simply multiply the number of kilojoules by 1000, since 1 kJ equals 1000 J.
With our example, the heat needed was 3.34 kJ, so:\[ 3.34 \text{ kJ} \times 1000 = 3340 \text{ J} \]Accurate conversions are key to correct results, ensuring data is consistent and aligns with expected scientific norms.

Joules to Kilojoules

Converting joules to kilojoules is a straightforward process. It's frequently needed when comparing or reporting energy values, especially in contexts where smaller units are cumbersome. Knowing that 1 kilojoule (kJ) is equivalent to 1000 joules (J) helps in making this conversion easy.
For instance, if you have 3340 J and want to convert it into kJ, simply divide by 1000:\[ 3340 \text{ J} \div 1000 = 3.34 \text{ kJ} \]Converting between these units helps in situations where you might have large scale energy applications and need a more concise representation of energy.