Question

If a 4.0 gram sample of iron metal absorbs 127.8 J of heat, causing its temperature to rise from 25°C to 97°C, what is the specific heat of iron?

Answer 1

To find the specific heat of iron, we can use the formula for heat transfer:

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

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°C),
- \( \Delta T \) is the change in temperature (in °C).

Given:
- \( q = 127.8 \, \text{J} \)
- \( m = 4.0 \, \text{g} \)
- Initial temperature = 25°C
- Final temperature = 97°C

First, calculate the change in temperature (\( \Delta T \)):

\[
\Delta T = \text{Final temperature} - \text{Initial temperature} = 97°C - 25°C = 72°C
\]

Now, we can rearrange the formula to solve for specific heat (\( c \)):

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

Substituting the known values into the equation:

\[
c = \frac{127.8 \, \text{J}}{4.0 \, \text{g} \cdot 72 \, \text{°C}}
\]

Calculating the denominator:

\[
4.0 \, \text{g} \cdot 72 \, \text{°C} = 288 \, \text{g°C}
\]

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

\[
c = \frac{127.8 \, \text{J}}{288 \, \text{g°C}}
\]

Calculating this gives:

\[
c \approx 0.4444 \, \text{J/g°C}
\]

Therefore, the specific heat of iron is approximately:

\[
\boxed{0.444 \, \text{J/g°C}}
\]