Final answer:
To heat 4.11 kg of water from 18.96°C to 34.40°C, the amount of heat required is 263.78 kJ, which is calculated using the specific heat capacity of water and the temperature change.
Step-by-step explanation:
To calculate the amount of heat required to heat 4.11 kg of water from 18.96°C to 34.40°C, we use the formula Q = m × c × ΔT, where Q is the heat transferred, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the temperature change. The specific heat capacity (c) for water is 4.184 J/g°C. To find Q, first convert the mass from kilograms to grams (4.11 kg = 4110 g), then calculate ΔT (34.40°C - 18.96°C = 15.44°C).
Now, substitute the values into the formula: Q = 4110 g × 4.184 J/g°C × 15.44°C, which gives Q = 263776.6784 J. Converting joules to kilojoules, this is approximately 263.78 kJ. Hence, Q = 263.78 kJ.