Question

The temperature of a 24.4 g sample of gold increases from 23.7°C to 30.8°C. If the specific heat of gold is 0.129 J/g-K, how many joules of heat are absorbed?

Answer 1

To calculate the amount of heat absorbed by the sample, you can use the formula for heat transfer, which is given by:

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

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

Given:
- Mass of gold \( m = 24.4 \, \text{g} \)
- Specific heat capacity of gold \( c = 0.129 \, \text{J/g-K} \)
- Initial temperature \( T_{\text{initial}} = 23.7 \, \text{°C} \)
- Final temperature \( T_{\text{final}} = 30.8 \, \text{°C} \)

First, we need to calculate the change in temperature:

\[
\Delta T = T_{\text{final}} - T_{\text{initial}} = 30.8 \, \text{°C} - 23.7 \, \text{°C} = 7.1 \, \text{°C}
\]

Now, we can plug the values into the heat transfer formula:

\[
q = (24.4 \, \text{g}) \cdot (0.129 \, \text{J/g-K}) \cdot (7.1 \, \text{K})
\]

Calculating this step by step:

1. Calculate \( (24.4 \, \text{g}) \cdot (0.129 \, \text{J/g-K}) \):
\[
24.4 \times 0.129 = 3.1506 \, \text{J/K}
\]

2. Now multiply by the change in temperature \( (7.1 \, \text{K}) \):
\[
q = 3.1506 \, \text{J/K} \cdot 7.1 \, \text{K} = 22.38926 \, \text{J}
\]

Rounding this to three significant figures (based on the precision of the given values), we find:

\[
q \approx 22.4 \, \text{J}
\]

Thus, the amount of heat absorbed by the gold is approximately **22.4 joules**.