Answer:
1.5 × 10³ g
Explanation:
Step 1: Given and required data
- Transferred heat (Q): 41,840 J
- Initial temperature: 22.0 °C
- Final temperature: 28.5 °C
- Specific heat capacity of water (c): 4.184 J/g.°C
Step 2: Calculate the temperature change
ΔT = 28.5°C - 22.0 °C = 6.5 °C
Step 3: Calculate the mass (m) of water
We will use the following expression.
Q = c × m × ΔT
m = Q / c × ΔT
m = 41,840 J / (4.184 J/g.°C) × 6.5 °C = 1.5 × 10³ g
