Question

A 80 gram piece of glass with the specific heat of 0.95 J/g°C is heated from 55°C to 77°C. How much heat is added to the glass? (1 point)

Item 26

5852 J

Item 21

4180 J

Item 28

1853 J

Item 29

1672 J

Answer 1

To calculate the amount of heat added to the glass, you can use the formula:

\[
Q = mc\Delta T
\]

where:
- \( Q \) is the heat added (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:
- Mass \( m = 80 \, \text{g} \)
- Specific heat \( c = 0.95 \, \text{J/g°C} \)
- Initial temperature \( T_i = 55\,°C \)
- Final temperature \( T_f = 77\,°C \)

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

\[
\Delta T = T_f - T_i = 77\,°C - 55\,°C = 22\,°C
\]

Now plug the values into the formula:

\[
Q = 80\, \text{g} \times 0.95\, \text{J/g°C} \times 22\,°C
\]

Calculating it step-by-step:

1. Calculate \( mc \):

\[
80\, \text{g} \times 0.95\, \text{J/g°C} = 76\, \text{J/°C}
\]

2. Now calculate \( Q \):

\[
Q = 76\, \text{J/°C} \times 22\,°C = 1672\, \text{J}
\]

Thus, the amount of heat added to the glass is:

\[
\boxed{1672 \, \text{J}}
\]