Physics 2019 JAMB Past Questions - Question 25: A 500W heater is used to heat 0.6kg of water from 250 C to 1000 C in t1 seconds. If another 1000W heater is used to heat 0.2kg of water from 100 C to 1000 C in t2 seconds, find t1 / t2

Correct Answer

D

Explanation

To solve this problem, we can use the formula:

Q = m * c * ΔT

Where:
Q = heat energy
m = mass of the water
c = specific heat capacity of water
ΔT = change in temperature

First, let's calculate the heat energy required to heat 0.6kg of water from 25°C to 100°C using the 500W heater.

Q1 = m * c * ΔT
  = 0.6kg * 4200J/kg°C * (100°C - 25°C)
  = 0.6kg * 4200J/kg°C * 75°C
  = 189,000J

Now, let's calculate the time taken (t1) using the 500W heater.

Power (P) = Energy / Time
500W = 189,000J / t1
t1 = 189,000J / 500W
t1 = 378 seconds

Next, let's calculate the heat energy required to heat 0.2kg of water from 100°C to 1000°C using the 1000W heater.

Q2 = m * c * ΔT
  = 0.2kg * 4200J/kg°C * (1000°C - 100°C)
  = 0.2kg * 4200J/kg°C * 900°C
  = 756,000J

Now, let's calculate the time taken (t2) using the 1000W heater.

Power (P) = Energy / Time
1000W = 756,000J / t2
t2 = 756,000J / 1000W
t2 = 756 seconds

Finally, we can find t1 / t2:

t1 / t2 = 378 seconds / 756 seconds
t1 / t2 = 0.5

So, the ratio of t1 to t2 is 0.5.