How much heat must be supplied to heat 100.3 g liquid mercury at 123.4°C to its boiling point at 356.7°C?

To calculate the amount of heat that must be supplied to heat liquid mercury from 123.4°C to its boiling point (356.7°C), we use the formula for heat transfer in a substance:

\[
Q = mc\Delta T
\]

Where:
- \( Q \) is the heat energy (in joules),
- \( m \) is the mass of the substance (in grams),
- \( c \) is the specific heat capacity of the substance (for mercury, \( c = 0.140 \ \text{J/g°C} \)),
- \( \Delta T \) is the change in temperature (final temperature - initial temperature).

### Step 1: Calculate \( \Delta T \)
\[
\Delta T = 356.7°C - 123.4°C = 233.3°C
\]

### Step 2: Apply the formula
\[
Q = 100.3 \ \text{g} \times 0.140 \ \text{J/g°C} \times 233.3°C
\]

Now let's compute the value.

\[
Q \approx 100.3 \times 0.140 \times 233.3 \ \text{J}
\] 

Let me calculate this.The amount of heat that must be supplied to heat 100.3 g of liquid mercury from 123.4°C to its boiling point at 356.7°C is approximately 3276 joules.

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:
1. How does the specific heat capacity of a substance affect the amount of heat required?
2. What is the heat capacity of other common substances like water or iron?
3. How do you calculate the heat needed for phase changes (e.g., melting or boiling)?
4. What role does mercury's low specific heat play in its temperature changes?
5. How would this calculation differ if the mercury were a solid or gas?

**Tip:** Always ensure units are consistent (e.g., mass in grams, temperature in °C) when performing heat-related calculations.

Calculate the heat required to raise 100.3 g of liquid mercury from 123.4°C to its boiling point at 356.7°C. This problem covers thermodynamics concepts, heat transfer, and the specific heat capacity formula. Suitable for students in Grades 9-12.

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