Question

2.

How many joules are liberated when 235 g of water are cooled from the boiling point to the
freezing point?

Answer 1

Q = 98030 joules

Step-by-step explanation:

To calculate the energy (in joules) liberated during the cooling of water, you can use the formula:

`\[ Q = m \cdot C \cdot \Delta T \]`

Where:

- Q is the heat energy,

- m is the mass of the substance (in grams),

- C is the specific heat capacity of the substance, and

-
`\( \Delta T \)` is the change in temperature.

For water, the specific heat capacity C is approximately
`\(4.18\ \text{J/g°C}\)`.

The change in temperature
`(\( \Delta T \))` is the difference between the initial and final temperatures. Assuming the boiling point of water is 100°C and the freezing point is 0°C,
`\( \Delta T = 100°C - 0°C = 100°C \)`.

Given that the mass
`(\( m \)) is \( 235\ \text{g} \)`, substitute these values into the formula:

`\[ Q = 235\ \text{g} * 4.18\ \text{J/g°C} * 100°C \]`

Q = 98030 joules

(Do not write the A's down I dont understand why they're there.)