Question

when 3.0g of water( with a mass of 4.179) is cooled from 80 Celsius to 10 Celsius. how much heat energy is lost?

Answer 1

To calculate the heat energy lost when cooling water, you can use the formula:

\[
q = mc\Delta T
\]

where:
- \( q \) is the heat energy lost (in joules),
- \( m \) is the mass of the substance (in grams),
- \( c \) is the specific heat capacity (for water, \( c = 4.179 \, \text{J/g}^\circ \text{C} \)),
- \( \Delta T \) is the change in temperature (in degrees Celsius).

### Given Data:
- Mass of water, \( m = 3.0 \, \text{g} \)
- Specific heat capacity of water, \( c = 4.179 \, \text{J/g}^\circ \text{C} \)
- Initial temperature, \( T_i = 80^\circ \text{C} \)
- Final temperature, \( T_f = 10^\circ \text{C} \)

### Calculating \(\Delta T\):
\[
\Delta T = T_f - T_i = 10^\circ \text{C} - 80^\circ \text{C} = -70^\circ \text{C}
\]

### Now, Substitute the values into the formula:
\[
q = mc\Delta T = (3.0 \, \text{g}) \times (4.179 \, \text{J/g}^\circ \text{C}) \times (-70^\circ \text{C})
\]

Calculating it:
\[
q = 3.0 \times 4.179 \times -70
\]
\[
q = 3.0 \times 4.179 = 12.537 \, \text{J}
\]
\[
q = 12.537 \times -70 = -877.59 \, \text{J}
\]

### Conclusion:
The heat energy lost by the water is approximately **877.59 J** (the negative sign indicates that heat is lost).