A 31.3 g sample of ethylene glycol, a car radiator coolant, loses 691 J of heat. What was the initial temperature if the final temperature is 32.5°C (c of ethylene glycol = 2.42 J/g·K)?

To solve for the initial temperature, we can use the specific heat equation:

\[
q = m \cdot c \cdot \Delta T
\]

Where:
- \( q \) is the heat lost (in joules, J),
- \( m \) is the mass (in grams, g),
- \( c \) is the specific heat capacity (in J/g·K),
- \( \Delta T \) is the change in temperature (in °C or K).

We are given:
- \( q = 691 \, \text{J} \),
- \( m = 31.3 \, \text{g} \),
- \( c = 2.42 \, \text{J/g·K} \),
- Final temperature \( T_f = 32.5 \, \degree\text{C} \).

### Step 1: Solve for \( \Delta T \)
Rearrange the formula to solve for \( \Delta T \):

\[
\Delta T = \frac{q}{m \cdot c}
\]

\[
\Delta T = \frac{691}{31.3 \cdot 2.42}
\]

### Step 2: Calculate \( \Delta T \)
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Would you like to calculate \( \Delta T \) yourself or ask for further assistance? Here's the next step:

Once you've found \( \Delta T \), the initial temperature can be calculated using:

\[
T_i = T_f + \Delta T
\]

Where:
- \( T_i \) is the initial temperature,
- \( T_f \) is the final temperature.

Do you have any questions or need more details on how to calculate it? Here are some related questions you might find helpful:

1. What does specific heat capacity depend on?
2. How can we measure heat transfer in real-world applications?
3. Why does ethylene glycol make an effective coolant?
4. How does temperature affect heat transfer?
5. What are the units of heat capacity?

**Tip:** Always ensure that the units for mass, specific heat, and temperature change are consistent when using the heat equation.

Solve for the initial temperature of a 31.3 g sample of ethylene glycol that loses 691 J of heat. This problem covers thermodynamic concepts like heat transfer, temperature change, and specific heat capacity. Detailed solution using q = m · c · ΔT formula. Suitable for high school students in Grades 9-12.

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