Question

The specific heat capacity of liquid water is 4.18 J/g-K. How many joules of heat are needed to raise the temperature of 6.00 g of water from 20.0 °C to 44.1

Answer 1

To calculate the amount of heat needed to raise the temperature of a substance, we can use the formula:

\[
q = m \cdot c \cdot \Delta T
\]

where:
- \( q \) = heat absorbed or released (in joules)
- \( m \) = mass of the substance (in grams)
- \( c \) = specific heat capacity (in J/g-K)
- \( \Delta T \) = change in temperature (in °C or K)

1. **Identify the values:**
- \( m = 6.00 \, \text{g} \)
- \( c = 4.18 \, \text{J/g-K} \)
- Initial temperature \( T_i = 20.0 \, \text{°C} \)
- Final temperature \( T_f = 44.1 \, \text{°C} \)

2. **Calculate the change in temperature \( \Delta T \):**
\[
\Delta T = T_f - T_i = 44.1 \, \text{°C} - 20.0 \, \text{°C} = 24.1 \, \text{°C}
\]

3. **Substitute the values into the formula:**
\[
q = 6.00 \, \text{g} \cdot 4.18 \, \text{J/g-K} \cdot 24.1 \, \text{K}
\]

4. **Calculate \( q \):**
\[
q = 6.00 \cdot 4.18 \cdot 24.1
\]
\[
q = 6.00 \cdot 4.18 \approx 25.08
\]
\[
q \approx 25.08 \cdot 24.1 \approx 604.41 \, \text{J}
\]

Thus, the amount of heat needed to raise the temperature of 6.00 g of water from 20.0 °C to 44.1 °C is approximately **604.4 joules**.