Answer:
Q = 5 kCal
Solution:
The equation used for this problem is as follow,
Q = m Cp ΔT
Where;
Q = Heat = ?
m = mass = 500 g
Cp = Specific Heat Capacity = 4.18 J.g⁻¹.°C⁻¹
ΔT = Change in Temperature = 30 °C - 20 °C = 10 °C
Putting values in eq. 1,
Q = 500 g × 4.18 J.g⁻¹.°C⁻¹ × 10 °C
Q = 20900 J
Q = 4.99 kCal ≈ 5 kCal
