how much energy is required to obtain superheated steam at 115 degrees celcius from 25 grams of water at 10 degrees celcius?
Q=mc(deltaT)
find Q
http://www.ausetute.com.au/heatcapa.html
To calculate the energy required to obtain superheated steam from water, we need to consider several steps. Let's break it down:
Step 1: Calculating the energy required to heat water to its boiling point.
The energy required to heat a substance can be calculated using the formula:
Q = mcΔT
Where:
Q = energy (in joules)
m = mass of the substance (in grams)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (in °C)
For the first step, we need to calculate the energy required to heat the water from 10°C to its boiling point, which is 100°C.
Q1 = m * c * ΔT1
Where:
m = 25 grams (given)
c = 4.18 J/g°C (specific heat capacity of water)
ΔT1 = (100°C - 10°C) = 90°C
Q1 = 25 g * 4.18 J/g°C * 90°C
Q1 = 9415 J
Step 2: Calculating the energy required to vaporize the water.
The energy required to vaporize water is given by the formula:
Q2 = mL
Where:
Q2 = energy (in joules)
m = mass of the substance (in grams)
L = heat of vaporization of water (2.26 × 10^6 J/kg or 2.26 × 10^3 J/g)
For the second step, we need to calculate the energy required to vaporize the water from its boiling point to superheated steam at 115°C.
Q2 = m * L
Where:
m = 25 grams (given)
L = 2.26 × 10^3 J/g (heat of vaporization)
Q2 = 25 g * 2.26 × 10^3 J/g
Q2 = 56500 J
Step 3: Summing up the energy required for all steps.
To obtain the total energy required to obtain superheated steam at 115°C, we need to add together the energies calculated in steps 1 and 2.
Total energy = Q1 + Q2
Total energy = 9415 J + 56500 J
Total energy = 65915 J
Therefore, the energy required to obtain superheated steam at 115°C from 25 grams of water at 10°C is approximately 65915 Joules.