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)
Responses

1672 J
1672 J

1853 J
1853 J

5852 J
5852 J

4180 J

Answer 1

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

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

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:
- \(m = 80\) g
- \(c = 0.95\) J/g°C
- Initial temperature = 55°C
- Final temperature = 77°C

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

\[
\Delta T = T_{final} - T_{initial} = 77°C - 55°C = 22°C
\]

Now substitute these values into the formula:

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

Calculating this step by step:

1. Calculate \(80 \cdot 0.95 = 76\) J.
2. Now multiply \(76 \, \text{J} \cdot 22 \, \text{°C} = 1672 \, \text{J}\).

Thus, the amount of heat added to the glass is \(1672 \, \text{J}\).

The correct response is **1672 J**.