Answer:
0.70 J/g.°C
Explanation:
Step 1: Given data
- Mass of graphite (m): 402 g
- Heat absorbed (Q): 1136 J
- Initial temperature: 26°C
- Specific heat of graphite (c): ?
Step 2: Calculate the specific heat of graphite
We will use the following expression.
Q = c × m × ΔT
c = Q / m × ΔT
c = 1136 J / 402 g × (30°C - 26°C)
c = 0.70 J/g.°C
The answer is 4.
Gases have low densities, because of the increased space between hight-energy particles.
Answer:
17
Explanation:
what is the independent variable
