90 calories of heat is added to 1 gram of fresh water with an initial temperature of 10 degrees Celcius. What is the final temperature of this liquid? (latent heat of fusion = 80cal/g°C, latent heat of vaporization = 540 cal/g°C)
q = mass water x specific heat H2O x (Tfinal-Tintial)
Substitute and solve for Tfinal.
To find the final temperature of the liquid, we need to calculate the amount of heat absorbed by the water.
First, we need to determine whether any phase changes occur during the process. Since the latent heat values for fusion and vaporization are given, we should check if the added heat is enough to cause a phase change.
The heat required to raise the temperature of 1 gram of water by 1 degree Celsius is known as the specific heat capacity (c) of water, which is approximately 1 cal/g°C.
The heat absorbed by water can be calculated using the formula:
Q = mcΔT
Where:
Q = heat absorbed
m = mass of the water (1 gram)
c = specific heat capacity of water (1 cal/g°C)
ΔT = change in temperature
Using the given data, we can find the change in temperature:
Q = 90 cal
m = 1 g
c = 1 cal/g°C
90 cal = (1 g)(1 cal/g°C)(ΔT)
ΔT = 90°C
Since the change in temperature is 90°C, we need to check whether any phase change occurs within this range.
The freezing point of water is 0°C, and the boiling point is 100°C.
If the change in temperature is within the range of 0°C to 100°C, no phase change occurs. Hence, the final temperature will be the initial temperature plus the change in temperature:
Final temperature = 10°C + 90°C = 100°C
Therefore, the final temperature of the liquid will be 100 degrees Celsius.