Question

when 3.0g of water 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:
- \( q \) = heat energy lost (in joules)
- \( m \) = mass of the water (in grams)
- \( c \) = specific heat capacity of water (approximately \( 4.18 \, \text{J/g}^\circ\text{C} \))
- \( \Delta T \) = change in temperature (in °C)

1. **Determine the mass of the water (m)**: \( 3.0 \, \text{g} \)
2. **Specific heat capacity of water (c)**: \( 4.18 \, \text{J/g}^\circ\text{C} \)
3. **Calculate the change in temperature (ΔT)**:
\[
\Delta T = T_{\text{final}} - T_{\text{initial}} = 10^\circ\text{C} - 80^\circ\text{C} = -70^\circ\text{C}
\]

4. **Now plug the values into the formula**:
\[
q = (3.0 \, \text{g}) \times (4.18 \, \text{J/g}^\circ\text{C}) \times (-70^\circ\text{C})
\]

Calculating:
\[
q = 3.0 \times 4.18 \times -70
\]
\[
q = 3.0 \times 4.18 \times -70 = -878.4 \, \text{J}
\]

Since we are interested in the heat lost, we take the absolute value:
\[
\text{Heat energy lost} = 878.4 \, \text{J}
\]

Therefore, the heat energy lost when 3.0 g of water is cooled from 80 °C to 10 °C is approximately **878.4 J**.