Question

A sample of iron weighing 111.6 g was subjected to heating, experiencing a temperature increase from 0°C to 20°C. The sample absorbed a total of 1004 J of energy during this process. What is the specific heat capacity of iron? (1 point)

0.45 310°C

4.18 Jig C

0.00578 Jig°C

2.20 Jig C

Answer 1

To calculate the specific heat capacity (c) of iron, we can use the formula:

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

Where:
- \( Q \) is the heat absorbed (in Joules),
- \( m \) is the mass (in grams),
- \( c \) is the specific heat capacity (in J/g°C),
- \( \Delta T \) is the change in temperature (in °C).

Given:
- \( Q = 1004 \) J,
- \( m = 111.6 \) g,
- \( \Delta T = 20°C - 0°C = 20°C \).

We need to rearrange the formula to solve for \( c \):

\[
c = \frac{Q}{m \cdot \Delta T}
\]

Substituting in the values:

\[
c = \frac{1004 \, \text{J}}{111.6 \, \text{g} \cdot 20 \, °C}
\]

Calculating the denominator:

\[
m \cdot \Delta T = 111.6 \, \text{g} \cdot 20 \, °C = 2232 \, \text{g°C}
\]

Now substitute this back into the equation for \( c \):

\[
c = \frac{1004 \, \text{J}}{2232 \, \text{g°C}}
\]

Calculating \( c \):

\[
c \approx 0.449 \text{ J/g°C}
\]

Rounding this value, we find:

\[
c \approx 0.45 \text{ J/g°C}
\]

Therefore, the specific heat capacity of iron is approximately **0.45 J/g°C**.