Answer :
Answer:
3.7 °C
Explanation:
Step 1: Given and required data
- Added heat (Q): 15 J
- Mass of water (m): 7.3 g
- Initial temperature (T₁): 3.2 °C
- Specific heat capacity of water (c): 4.184 J/g.°C
Step 2: Calculate the final temperature (T₂) of the water
We will use the following expression.
Q = c ×m ×(T₂ - T₁)
T₂ = Q / c × m + T₁
T₂ = 15 J / (4.184 J/g.°C) × 7.3 g + 3.2 °C = 3.7 °C