Navjyot Kalra
Last Activity: 10 Years ago
To solve this problem, we will use the formula for heat energy:
Q = mcΔT
where:
Q = heat energy (Joules)
m = mass of water (grams)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = temperature change (final temperature - initial temperature)
Given:
m = 136 g
c = 4.18 J/g°C
Initial temperature = 23.5°C
Final temperature = 100°C (boiling point of water)
Power of heater = 220 W (1 W = 1 J/s)
Step 1: Calculate the required heat energy (Q)
ΔT = 100 - 23.5 = 76.5°C
Q = (136 g) × (4.18 J/g°C) × (76.5°C)
Q = 43491.72 J
Step 2: Calculate time required using power equation
Power = Energy / Time
Time = Energy / Power
Time = 43491.72 J / 220 J/s
Time ≈ 197.7 seconds
Final Answer: The time required to bring the water to a boil is approximately 198 seconds (or about 3 minutes and 18 seconds).