A large pot is placed on a stove and 1.2 kg of water at 14°C is added to the pot. The temperature of the water is raised evenly to 100°C just before it starts to boil. (a) What amount of heat is absorbed by the water in reaching 100°C? (b) The water then boils until all of it has evaporated, turning to water vapor at 100°C. How much heat does the water absorb in this process?
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To find the amount of heat absorbed by the water in the scenarios described, we can use the equation:
Q = mcΔT
Where:
Q is the heat absorbed (in joules),
m is the mass of the substance (in kilograms),
c is the specific heat capacity of the substance (in joules per kilogram per degree Celsius), and
ΔT is the change in temperature (in degrees Celsius).
(a) Finding the heat absorbed in raising the temperature:
First, let's calculate the mass of the water:
m = 1.2 kg
Next, we need to find the change in temperature:
ΔT = (final temperature - initial temperature)
Given that the initial temperature is 14°C and the final temperature is 100°C, we have:
ΔT = 100°C - 14°C = 86°C
Now, we need to utilize the specific heat capacity of water, which is approximately 4186 J/kg°C.
Plugging the values into the formula, we get:
Q = (1.2 kg) * (4186 J/kg°C) * (86°C)
Q = 439,149.6 J
Therefore, the amount of heat absorbed by the water in reaching 100°C is approximately 439,150 J.
(b) Finding the heat absorbed during boiling:
In this scenario, the water remains at 100°C, but undergoes a phase change from a liquid to a vapor. During boiling, the heat absorbed is related to the latent heat of vaporization.
The specific latent heat of vaporization of water is approximately 2,260,000 J/kg.
To find the heat absorbed during boiling, we need to multiply the mass of the water by the specific latent heat of vaporization:
Q = (1.2 kg) * (2,260,000 J/kg)
Q = 2,712,00 J
Therefore, the amount of heat absorbed by the water during boiling is approximately 2,712,00 J.