Answer:
the energy removed when cooling the water is 602304J.
Explanation:
To calculate the energy removed when cooling water from steam at 125°C to 53°C, we can use the specific heat capacity formula:
Q=mcΔT
Where:
Q is the energy removed (in joules),
m is the mass of the water (in kilograms),
c is the specific heat capacity of water (in joules per kilogram per degree Celsius),
ΔT is the change in temperature (in degrees Celsius).
Given:
m=0.2 kg (mass of water),
c=4186 J/kg°C (specific heat capacity of water),
T₁ =125°C (initial temperature of the water),
T₂ =53°C (final temperature of the water).
First, we need to calculate the temperature change:
ΔT=T₂ − T₁ =53°C−125°C=−72°C
Now, we can substitute the values into the specific heat capacity formula:
Q=(0.2kg)(4186J/kg°C)(−72°C)
Q=−602304J
Since the temperature change is negative, indicating a decrease in temperature, the energy removed is positive. Therefore, the energy removed when cooling the water is 602304J.
