A 100.0g sample of water at 20.0C is heated to steam at 125.0c.how many heat was absorbed?
Add these
a. Heat to heat water=100*cwater*(100-20)
b. heat to vaporize=100*HeatVaporization
c. heat to heat steam=100*csteam(125-100)
look up the heat of vaporization of water, the specific heat of water, and the specific heat of steam. Watch units, mass is in grams
Well, let me put on my detective cap and investigate this heating situation.
So, to find out how much heat was absorbed, we need to use the equation Q = m * c * ΔT. Here, Q is the heat absorbed, m is the mass of the water, c is the specific heat of water, and ΔT is the change in temperature.
First, let's find the mass of the water. It's given as 100.0g, so we'll use that.
Secondly, we need to know the specific heat of water, which is approximately 4.18 J/g°C.
Lastly, we calculate ΔT by subtracting the initial temperature from the final temperature: ΔT = (125.0°C - 20.0°C) = 105.0°C.
Putting it all together, we can calculate the heat absorbed:
Q = 100.0g * 4.18 J/g°C * 105.0°C
Now, let me use my amazing calculating powers.
*Clown Bot activates mental calculator*
Drumroll, please...
After crunching some numbers, the heat absorbed is approximately 438,900 Joules. That's a lot of heat!
To calculate the amount of heat absorbed when a substance undergoes a temperature change, you can use the equation:
q = m * C * ΔT
Where:
q = heat absorbed/lost (in Joules)
m = mass of the substance (in grams)
C = specific heat capacity of the substance (in J/g°C)
ΔT = change in temperature (in °C)
In this case, we are dealing with water, and the specific heat capacity of water is 4.18 J/g°C.
Given:
m = 100.0 g
C = 4.18 J/g°C
Initial temperature (T1) = 20.0°C
Final temperature (T2) = 125.0°C
First, let's calculate the change in temperature (ΔT):
ΔT = T2 - T1
ΔT = 125.0°C - 20.0°C
ΔT = 105.0°C
Now, we can calculate the amount of heat absorbed (q):
q = m * C * ΔT
q = 100.0 g * 4.18 J/g°C * 105.0°C
q = 438,900 J
Therefore, approximately 438,900 Joules of heat were absorbed.
To find out how much heat was absorbed in this process, we can use the equation:
q = m * ΔT * C
where:
q = heat absorbed/lost (in joules)
m = mass of the substance (in grams)
ΔT = change in temperature (in degrees Celsius)
C = specific heat capacity of the substance (in J/g°C)
For water, the specific heat capacity is 4.18 J/g°C.
Given:
m = 100.0g
ΔT = (125.0°C - 20.0°C) = 105.0°C
Substituting the values into the equation:
q = 100.0g * 105.0°C * 4.18 J/g°C
= 439,350 J
Therefore, the amount of heat absorbed is 439,350 joules.