Question

1. how much energy is released by 650g of water as it cools from its boiling point to its freezing point? the specific heat capacity of water is 4.18 j/g·°c. record your answer to the nearest kilojoules. enter your answer in the box provided. kilojoules

Explanation:

Step1: Identify the formula for heat energy

The formula for heat energy \( Q \) is \( Q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature.

Step2: Determine the values

• Mass \( m = 650 \, \text{g} \)
• Specific heat capacity \( c = 4.18 \, \text{J/g}^\circ\text{C} \)
• Change in temperature \( \Delta T \): Boiling point of water is \( 100^\circ\text{C} \), freezing point is \( 0^\circ\text{C} \), so \( \Delta T = 100 - 0 = 100^\circ\text{C} \)

Step3: Calculate the heat energy in joules

Substitute the values into the formula:
\( Q = 650 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C} \times 100^\circ\text{C} \)
\( Q = 650 \times 4.18 \times 100 \)
\( Q = 650 \times 418 \)
\( Q = 271700 \, \text{J} \)

Step4: Convert joules to kilojoules

Since \( 1 \, \text{kJ} = 1000 \, \text{J} \), divide by 1000:
\( Q = \frac{271700}{1000} = 271.7 \, \text{kJ} \)

Answer:

271.7