Heat totaling 2700.0 J leaves a glass of water at 75.0 C to bring it to room temperature (20.0 C). What is the mass of the water in g (ignore the glass)?
2700 = m *specific heat of water * (75-20)
so you find the specific heat of water, then multiply that times the change in temperature then divide that into the joules?
Yes
about 4180 Joules/ kg deg K
as I remember
By the way change in degrees K is the same as change in degrees C
C = 273 + K
so delta C = delta K
To find the mass of the water in grams, we need to use the specific heat capacity equation, which is:
Q = m * c * ΔT
Where:
Q is the heat added or removed (in Joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g·°C)
ΔT is the change in temperature (in °C)
In this case, we know:
Q = -2700.0 J (negative since heat is leaving the water)
ΔT = (20.0 - 75.0) °C = -55.0 °C (negative since the temperature is decreasing)
c = 4.184 J/g·°C (specific heat capacity of water at approximately 20 °C)
Now we can rearrange the equation to solve for mass (m):
m = Q / (c * ΔT)
Substituting the given values:
m = -2700.0 J / (4.184 J/g·°C * -55.0 °C)
Calculating this expression will give us the mass of the water in grams. Let's plug these numbers into a calculator.