A 152 g sample of ice at –37 degree C is heated until it turns into liquid water at 0 degree C. How do you find the change in heat content in the system?
I've found H(I) [Change in heat of the ice] to be 11416.72 but im still stuck on the change of heat for the water.
The change in heat content in the system can be calculated using the specific heat capacity of water. The specific heat capacity of water is 4.184 J/g°C. Therefore, the change in heat content in the system is:
Change in heat content = (152 g)(4.184 J/g°C)(0°C - (-37°C)) = 11416.72 J
To find the change in heat content for the water, you can use the equation:
Q = m * c * ΔT
Where:
Q is the change in heat content (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in joules per gram per degree Celsius),
ΔT is the change in temperature (in degrees Celsius).
For water, the specific heat capacity (c) is approximately 4.18 J/g°C.
In this case, the mass (m) of the water is equal to the mass of the ice, which is 152 g. The change in temperature (ΔT) is the difference between the final temperature (0°C) and the initial temperature (-37°C), which is 37°C.
Substituting the values into the equation, we have:
Q = 152 g * 4.18 J/g°C * 37°C
Calculating this, you will find the change in heat content (Q) for the water.
To find the change in heat content for the water, you can use the equation:
Q = m * c * ΔT
Where:
Q is the change in heat content
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
Since the ice at -37 degrees Celsius is heated until it becomes liquid water at 0 degrees Celsius, ΔT would be 0 degrees Celsius - (-37 degrees Celsius) = 37 degrees Celsius.
The mass of the water can be calculated using the mass of the ice. The assumption here is that the mass of the ice remains constant throughout the phase change.
mass of water = mass of ice = 152 g
The specific heat capacity of water is 4.18 J/g°C.
Now, you can substitute these values into the equation and calculate the change in heat content for the water:
Q = (152 g) * (4.18 J/g°C) * (37°C)
Q = 22471.04 J
Therefore, the change in heat content for the water is 22471.04 J.