How many joules are required to heat 200. grams of water from 25 °C to 125°C?

To calculate the amount of energy in joules required to heat water, we can use the formula:

\[
q = mc\Delta T
\]

Where:
- \(q\) is the heat energy in joules.
- \(m\) is the mass of the water in grams (200 g).
- \(c\) is the specific heat capacity of water, which is \(4.18 \, \text{J/g}^\circ\text{C}\).
- \(\Delta T\) is the change in temperature (final temperature - initial temperature).

Since the final temperature is 125°C and the initial temperature is 25°C, we have:

\[
\Delta T = 125^\circ\text{C} - 25^\circ\text{C} = 100^\circ\text{C}
\]

Now plug the values into the equation:

\[
q = (200 \, \text{g}) \times (4.18 \, \text{J/g}^\circ\text{C}) \times (100^\circ\text{C})
\]

Let me calculate that.

\[
q = 200 \times 4.18 \times 100 = 83,600 \, \text{J}
\]

So, **83,600 joules** are required to heat 200 g of water from 25°C to 125°C.

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. How would the energy required change if the specific heat capacity of water was lower?
2. How much energy is required to heat the same amount of water from 0°C to 100°C?
3. What is the relationship between temperature change and heat energy?
4. How much energy is needed to boil 200 g of water at 100°C into steam?
5. What role does heat energy play in phase changes like melting or boiling?

**Tip**: Always check if the temperature exceeds water's boiling point (100°C) to account for phase change energy, such as latent heat.

Find out how to calculate the energy required to heat 200 grams of water from 25°C to 125°C using the specific heat capacity of water. This step-by-step solution covers thermodynamic principles and the heat energy formula q = mcΔT. Suitable for students in Grades 9-11.

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