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?
A. 0.044 J/g°C
B. 0.444 J/g°C
C. 0.0044 J/g°C
D. 4.4 J/g°C

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 (in grams),
- \( c \) is the specific heat capacity (in J/g°C),
- \( \Delta T \) is the change in temperature (in °C).

In this case:
- \( q = 127.8 \) J,
- \( m = 4.0 \) g,
- Initial temperature = 25°C,
- Final temperature = 97°C.

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

\[
\Delta T = 97°C - 25°C = 72°C
\]

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

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

Plugging in the values:

\[
c = \frac{127.8 \, \text{J}}{4.0 \, \text{g} \cdot 72 \, \text{°C}} = \frac{127.8}{288} \approx 0.444 \, \text{J/g°C}
\]

Thus, the specific heat of iron is approximately \( 0.444 \, \text{J/g°C} \).

The correct answer is:
**B. 0.444 J/g°C**.