Explanation:
We need to go through to stages to boil 41.1 g of water. We have to heat the sample of water from 84.7 °C to 100 °C (the boiling point) And then we have to provide enough heat to boil all the sample of water.
<em>a) Heating from 84.7 °C to 100 °C:</em>
This is calculated using the formula:
Q₁ = m * C * ΔT
Where Q₁ is the amount of heat, m is the mass of the sample, C is the specific heat of water and ΔT is the temperature change. We already know these values:
m = 41.1 g
C = 4.184 J/(g*°C)
ΔT = Tfinal - Tinitial = 100 °C - 84.7 °C
ΔT = 15.3 °C
Replacing these values we can get the amount of heat necessary for the first step:
Q₁ = m * C * ΔT
Q₁ = 41.1 g * 4.184 J/(g°C) * 15.3 °C
Q₁ = 2631 J
<em>b) Boiling 41.1 g of water:</em>
To find the amount of heat that we need to provide to the sample of water to completely boil it we can use this formula:
Q₂ = m * Cv
Where Cv is the latent heat of vaporization.
Cv = 2256 J/g
Q₂ = m * Cv
Q₂ = 41.1 g * 2256 J/g
Q₂ = 92721 J
<em>c) Total amount of heat:</em>
Qtotal = Q₁ + Q₂
Qtotal = 2631 J + 92721 J
Qtotal = 95352 J = 95400 J
Qtotal = 95.4 kJ
Answer: The amount of heat needed to boil the sample of water is 95.4 kJ or 95400 J.
