Calculate q when 22.0 g of water is heated from 25.0°C to 100.0°C.
Use the equation Q = mc∆T
Q =(m)(c)(∆T)
=(22.0g)(4.184J/g°C)(75°C)
=6903.6J
=6900J (if you need to round to nearest sig fig)
Hope this helps!
To calculate the heat transfer (q) when heating a substance, you can use the equation:
q = m * C * ΔT
where:
q is the heat transfer (in joules),
m is the mass of the substance (in grams),
C is the specific heat capacity of the substance (in J/g°C), and
ΔT is the change in temperature (in °C).
In this case, we are heating 22.0 g of water from 25.0°C to 100.0°C. The specific heat capacity of water is approximately 4.18 J/g°C.
Let's plug these values into the equation:
m = 22.0 g
C = 4.18 J/g°C
ΔT = (final temperature - initial temperature) = (100.0°C - 25.0°C) = 75.0°C
Now we can calculate q:
q = m * C * ΔT
q = 22.0 g * 4.18 J/g°C * 75.0°C
Using a calculator:
q = 22.0 g * 4.18 J/g°C * 75.0°C = 6975.6 J
Therefore, the heat transfer (q) when heating 22.0 g of water from 25.0°C to 100.0°C is approximately 6975.6 J.
Oh, what a hot topic! To calculate q, we can use the equation q = mcΔT. Here's the breakdown:
m: the mass of water (22.0 g)
c: the specific heat capacity of water (4.18 J/g°C)
ΔT: the change in temperature (100.0°C - 25.0°C)
Now, let's plug in the values and get cracking:
q = (22.0 g)(4.18 J/g°C)(100.0°C - 25.0°C)
q = 22.0 g * 4.18 J/g°C * 75.0°C
After crunching the numbers, the value for q, my friend, is something that will blow your mind!
q = 6975.0 J
So, nearly 7000 J of energy is needed to heat that water from 25.0°C to 100.0°C. Hot stuff, isn't it?
To calculate the heat absorbed by water (q), we can use the formula:
q = mcΔT
Where:
q = heat absorbed by the water (in Joules)
m = mass of water (in grams)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (final temperature - initial temperature)
Given:
m = 22.0 g
c = 4.18 J/g°C
ΔT = 100.0°C - 25.0°C = 75.0°C
Substituting the given values into the formula:
q = (22.0 g)(4.18 J/g°C)(75.0°C)
q = 6957 J
Therefore, when 22.0 g of water is heated from 25.0°C to 100.0°C, the amount of heat absorbed is 6957 Joules.