Energy in the amount of 420J is added to a 35g sample of water at a temperature of 10.0 C . What is the final temperature of water ?
420=35g*cwater*(tf-10)
put cwater in (watch units, need J/gC)
solve for tf
To determine the final temperature of the water, we need to use the formula:
Q = mcΔT
Where:
Q is the heat energy added (420J),
m is the mass of water (35g),
c is the specific heat capacity of water (4.18 J/g°C),
and ΔT is the change in temperature.
First, we rearrange the formula to solve for ΔT:
ΔT = Q / (mc)
Let's substitute the given values into the formula:
ΔT = 420J / (35g * 4.18 J/g°C)
Calculating:
ΔT = 420J / 146.3J/°C
ΔT ≈ 2.87°C
Now, we can find the final temperature by adding the change in temperature (ΔT) to the initial temperature (10.0°C):
Final temperature = 10.0°C + 2.87°C
Therefore, the final temperature of the water is approximately 12.87°C.
To determine the final temperature of the water, we can use the equation:
Q = mcΔT
Where:
Q is the heat energy transferred to the water
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
First, let's rearrange the equation to solve for ΔT:
ΔT = Q / (mc)
Now we can substitute the given values into the equation:
Q = 420 J
m = 35 g
c = 4.18 J/g°C (specific heat capacity of water)
ΔT = 420 J / (35 g x 4.18 J/g°C)
Calculating this expression will give us the change in temperature. We need to remember that the final temperature will be the initial temperature (10.0°C) plus the change in temperature.
Let's calculate:
ΔT = 420 J / (35 g x 4.18 J/g°C)
ΔT ≈ 3.77°C
Therefore, the final temperature of the water will be the initial temperature (10.0°C) plus the change in temperature (3.77°C):
Final temperature = 10.0°C + 3.77°C
Final temperature ≈ 13.77°C
Hence, the final temperature of the water will be approximately 13.77°C.