Question

If 127.8 J of heat is added to a 5.0 gram sample of substance X metal and the temperature of the metal increases from 25°C to 85°C, what is the specific heat of substance X?

Answer 1

To calculate the specific heat of substance X, we can use the formula for heat transfer:

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

where:
- \( q \) is the heat added (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 the data:
- \( q = 127.8 \, \text{J} \)
- \( m = 5.0 \, \text{g} \)
- Initial temperature = 25°C
- Final temperature = 85°C

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

\[
\Delta T = \text{Final Temperature} - \text{Initial Temperature} = 85°C - 25°C = 60°C
\]

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

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

Substituting in the known values:

\[
c = \frac{127.8 \, \text{J}}{5.0 \, \text{g} \cdot 60°C}
\]

Now calculate the denominator:

\[
m \cdot \Delta T = 5.0 \, \text{g} \cdot 60°C = 300 \, \text{g°C}
\]

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

\[
c = \frac{127.8 \, \text{J}}{300 \, \text{g°C}} \approx 0.426 \, \text{J/g°C}
\]

Thus, the specific heat of substance X is approximately:

\[
\boxed{0.426 \, \text{J/g°C}}
\]