A 1 g sample of water at 373 K is cooled by the removal of 100 J . what will be the final temperature of the water
See your other post above. Same idea.
To find the final temperature of the water, we can use the equation:
Q = mcΔT
Where:
Q is the amount of heat transferred
m is the mass of the substance (in this case, water)
c is the specific heat capacity of water
ΔT is the change in temperature
In this case, we are given that the initial temperature is 373 K and the heat removed (Q) is 100 J. We need to find the final temperature. Given that the mass of the water is 1 g, we also need to know the specific heat capacity of water.
The specific heat capacity of water is approximately 4.18 J/g°C. This means that it takes 4.18 joules of heat to raise the temperature of 1 gram of water by 1 degree Celsius.
Now, let's substitute the given values into the equation and find the final temperature:
Q = mcΔT
100 J = (1 g) * (4.18 J/g°C) * (Tf - 373 K)
Dividing both sides of the equation by (1 g) * (4.18 J/g°C), we get:
100 J / [(1 g) * (4.18 J/g°C)] = Tf - 373 K
Now, let's solve for Tf:
Tf = 100 J / [(1 g) * (4.18 J/g°C)] + 373 K
Calculating this, we find:
Tf ≈ 100 / 4.18 + 373 ≈ 24 + 373 ≈ 397 K
Therefore, the final temperature of the water would be approximately 397 K.