Question

The temperature of a 6.0 g sample of glass changed from 20.0°C to 45.0°C when it absorbed 550 J of heat. What is the specific heat of this glass?

Answer 1

To calculate the specific heat capacity of the glass, 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:
- \( q = 550 \, \text{J} \),
- \( m = 6.0 \, \text{g} \),
- Initial temperature = \( 20.0 \, \text{°C} \),
- Final temperature = \( 45.0 \, \text{°C} \).

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

\[
\Delta T = \text{Final temperature} - \text{Initial temperature} = 45.0 \, \text{°C} - 20.0 \, \text{°C} = 25.0 \, \text{°C}
\]

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

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

Substituting the known values:

\[
c = \frac{550 \, \text{J}}{6.0 \, \text{g} \cdot 25.0 \, \text{°C}}
\]

Calculating the denominator:

\[
6.0 \, \text{g} \cdot 25.0 \, \text{°C} = 150 \, \text{g°C}
\]

Now, substituting back into the equation for specific heat:

\[
c = \frac{550 \, \text{J}}{150 \, \text{g°C}} = \frac{550}{150}
\]

Calculating the value:

\[
c = 3.67 \, \text{J/g°C}
\]

Thus, the specific heat of the glass is approximately:

\[
\boxed{3.67 \, \text{J/g°C}}
\]