If 395 grams of water at 21 Celsius absorbs 285kJ of heat from a mountain climbers stove at an elevation where the boiling point of water is 87 Celsius, is this amount of energy sufficient to heat the water to its boiling point? How many kJ are required to boil the water?
q needed to raise T of water from 21 C to 87 C is
q = mass water x specific heat water x (Tfinal-Tinitial).
I get approximately 110 kJ. Is that enough?
To determine if the amount of energy is sufficient to heat the water to its boiling point, we need to calculate the amount of energy required to raise the temperature of the water from 21 degrees Celsius to 87 degrees Celsius.
The energy required to heat a substance can be calculated using the formula:
Q = mcΔT
Where Q is the energy required, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
First, let's calculate the energy required to raise the temperature of the water from 21 degrees Celsius to 87 degrees Celsius:
Q = (395g)(4.18 J/g°C)(87 - 21)
Here, we are assuming the specific heat capacity of water is 4.18 J/g°C.
Converting grams to kilograms:
Q = (0.395kg)(4.18 J/g°C)(87 - 21)
Now, let's calculate the value of Q:
Q = (0.395kg)(4.18 J/g°C)(66)
Q = 109.5516 J
Converting J to kJ:
Q = 0.1096 kJ
So, it requires 0.1096 kJ of energy to heat the water from 21 degrees Celsius to 87 degrees Celsius.
Since the stove provides 285 kJ of heat, it is indeed sufficient to heat the water to its boiling point.
To determine the amount of energy required to boil the water, we need to calculate the specific heat of water during the phase change from liquid to vapor. This is known as the heat of vaporization.
The heat of vaporization of water is approximately 2260 kJ/kg.
To calculate the energy required to boil the water, we use the formula:
Q = mL
Where Q is the energy required, m is the mass of the water, and L is the heat of vaporization.
Let's calculate the energy required:
Q = (395g)(2260 J/g)
Converting grams to kilograms:
Q = (0.395kg)(2260 J/g)
Q = 892.7 J
Converting J to kJ:
Q = 0.8927 kJ
So, it requires 0.8927 kJ of energy to boil the water.
Therefore, the amount of energy provided by the stove is sufficient to heat the water to its boiling point, and an additional 0.8927 kJ is required to boil the water.