Answer:
8.55 × 10³ cal
Explanation:
Step 1: Given and required data
- Specific heat of water (c): 1 cal/g.°C
- Initial temperature: 22.7 °C
- Final temperature: 38.8 °C
Step 2: Calculate the temperature change (ΔT)
ΔT = Final temperature - Initial temperature = 38.8 °C - 22.7 °C = 16.1 °C
Step 3: Calculate the heat required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 1 cal/g.°C × 531 g × 16.1 °C = 8.55 × 10³ cal
